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37. (10-11 grades) How many planes are equidistant from four points that do not lie in the same plane?
|
37. Taking these points as the vertices of a tetrahedron, it is easy to establish that only seven planes can be drawn equidistant from its vertices.
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
64. (9th grade) The number 7 is raised to the seventh power, the resulting number is again raised to the seventh power, and so on. This process is repeated 1000 times. What is the last digit of this number?
|
64. Considering the powers of the number 7 in sequence, we notice that the last digits of these powers repeat every four, so the number given in the problem ends with the digit 7.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
81. The difference $\sqrt{|40 \sqrt{2}-57|}-\sqrt{40 \sqrt{2}+57}$ is an integer. Find this number.
|
Solution. Since $40 \sqrt{2}-57<0$, then $|40 \sqrt{2}-57|=$ $=57-40 \sqrt{2}$. Then
$$
\begin{aligned}
A & =\sqrt{|40 \sqrt{2}-57|}-\sqrt{40 \sqrt{2}+57}= \\
& =\sqrt{57-40 \sqrt{2}}-\sqrt{57+40 \sqrt{2}}
\end{aligned}
$$
Let $57-40 \sqrt{2}=(a+b \sqrt{2})^{2}$, where $a$ and $b$ are unknown coefficients. Then
$$
57-40 \sqrt{2}=a^{2}+2 b^{2}+2 \sqrt{2} a b
$$
from which
$$
\left\{\begin{array}{l}
a^{2}+2 b^{2}=57 \\
2 a b=-40
\end{array}\right.
$$
Solving this system of equations, we get $a=5, b=-4$. Therefore, $\sqrt{57-40 \sqrt{2}}=\sqrt{(5-4 \sqrt{2})^{2}}=|5-4 \sqrt{2}|=4 \sqrt{2}-5, \quad$ since $5-4 \sqrt{2}<0$.
Similarly, it is established that $\sqrt{57+40 \sqrt{2}}=4 \sqrt{2}+5$. Consequently, $A=4 \sqrt{2}-5-(4 \sqrt{2}+5)=-10$.
## Elimination of Irrationality in the Denominator
|
-10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
89. Simplify the expression:
a) $\sqrt{|40 \sqrt{2}-57|}-\sqrt{40 \sqrt{2}+57}$
b) $\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$
|
Solution. a) Let $\sqrt{57-40 \sqrt{2}}-\sqrt{57+40 \sqrt{\overline{2}}}=x ; x<0$, since $\sqrt{57-40 \sqrt{2}}<\sqrt{57+40 \sqrt{2}}$.
Square both sides of the equation:
$$
x^{2}=57-40 \sqrt{2}-2 \sqrt{57^{2}-(40 \sqrt{2})^{2}}+57+40 \sqrt{2}
$$
from which
$$
x^{2}=114-2 \sqrt{49}, x^{2}=100, x=-10
$$
(Solution by the method of undetermined coefficients see p. 51.)
b) Let the value of the expression be $x$, i.e.
$$
\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=x
$$
Cube both sides of the equation:
$$
\begin{aligned}
& 4+3 \cdot(\sqrt[3]{2+\sqrt{5}})^{2} \cdot \sqrt[3]{2-\sqrt{5}}+ \\
& +3 \cdot \sqrt[3]{2+\sqrt{5}} \cdot(\sqrt[3]{2-\sqrt{5}})^{2}=x^{3}
\end{aligned}
$$
or
$$
3 \sqrt[3]{2+\sqrt{5}} \cdot \sqrt[3]{2-\sqrt{5}} \cdot(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}})=x^{3}-4
$$
or (see (1))
$$
\begin{aligned}
3 \cdot \sqrt[3]{2^{2}-(\sqrt{5})^{2}} \cdot x & =x^{3}-4, \text { i.e. } \\
x^{3}+3 x-4 & =0
\end{aligned}
$$
Thus, the value $x$ of the expression is a root of equation (2).
It is obvious that the number $x=1$ is a root of equation (2).
|
-10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
90. Prove that $\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}=3$.
|
Solution. Let the value of the expression on the left side of the equality be denoted by $x$. Reasoning as in the solution of Example 89, b, we obtain the equation $x^{3}-3 x-18=0$, from which $x=3$.
## Solving Equations
|
3
|
Algebra
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
91. Let's solve the equation
$$
(3 x-1) \sqrt{x-4}=17 \sqrt{2}
$$
|
Solution. Method 1. The domain of expression (3) is the interval $[4 ;+\infty)$. Squaring both sides of the equation, we obtain the equivalent equation
$$
9 x^{3}-42 x^{2}+25 x-582=0
$$
We find the critical points of the function $f(x)=9 x^{3}-42 x^{2}+25 x-582$:
$$
f^{\prime}(x)=27 x^{2}-84 x+25=0
$$
from which $x=\frac{1}{3}$ and $x=\frac{25}{9}$. Both critical points do not belong to the domain of equation (3). Therefore, we look for the zero of the function in the interval $[4 ;+\infty)$, where $f^{\prime}(x)>0$. In this interval, the function is increasing, and it has a unique zero. Since
$$
\begin{aligned}
& f(4)=9 \cdot 4^{3}-42 \cdot 4^{2}+25 \cdot 4-5820
\end{aligned}
$$
by the Bolzano-Cauchy theorem, the zero of the function belongs to the interval $[4 ; 10]$. By trial, we find the root of the equation—it is equal to 6.
Method 2. Suppose that the factors containing radicals are equal. Then the system holds
$$
\left\{\begin{array}{l}
3 x-1=17 \\
\sqrt{x-4}=\sqrt{2}
\end{array}\right.
$$
Solving it, we get $x=6$. It is easy to verify that the number 6 is a root of the given equation. We will prove that the equation has no other roots.
Indeed, for $x>6$, the system of inequalities holds
$$
\left\{\begin{array}{l}
3 x-1>17 \\
\sqrt{x-4}>\sqrt{2}
\end{array}\right.
$$
Then
$$
(3 x-1) \cdot \sqrt{x-4}>17 \sqrt{2}
$$
If $4 \leqslant x<6$, then the system of inequalities holds
$$
\left\{\begin{array}{l}
3 x-1<17 \\
\sqrt{x-4}<\sqrt{2}
\end{array}\right.
$$
i.e.
$$
(3 x-1) \cdot \sqrt{x-4}<17 \sqrt{2}
$$
Therefore, the original equation has a unique root $x=6$.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
31. Cube. Holding a model of a cube in your hand so that it can rotate around its longest axis (i.e., around the line connecting opposite vertices), you can wind black yarn around it without any gaps. The yarn will shade only half of the cube (why?). The same can be done with another axis; there are four in total, and each time we use a different color of yarn (black, red, blue, and
yellow). The entire model will be covered with different colors, and from their mixing, mixed colors will emerge (the model of the cube is white and we do not consider this color). How many color shades will there be on the cube and which ones?
|
31. The yarn wound around a cube rotating about one of its axes (Fig. 41) will remain only on those edges that do not have common points with the axis of rotation; the yarn will cover half of each face of the cube, i.e., half of the cube's surface.
Now, let's rotate the cube sequentially around each of the four axes, each time winding yarn of a different color.
Thus, when rotating around the axis $A C^{\prime}$, we will use yarn of color $a$ (black); when rotating around the axis $D B^{\prime}$, we will use yarn of color $b$ (red); when rotating around the axis $B D^{\prime}$, we will use yarn of color $c$ (yellow); and finally, when rotating around the axis $C A^{\prime}$, we will use yarn of color $d$ (blue).
The cube will be colored as shown in Fig. 42, where each face is divided into four triangles; the letters placed in the corresponding triangles denote the colors of the yarn covering that triangle.
It is easy to notice that:
$1^{\circ}$ six shades will appear on the surface of the cube, i.e., as many as there are combinations of four elements taken two at a time, namely: $a b, a c, a d, b c, b d$, and $c d$;
$2^{\circ}$ each face will have four different shades;
$3^{\circ}$ the surface of the cube will be covered by two layers of yarn;
$4^{\circ}$ opposite faces of the cube will be colored in the same shades, arranged in reverse cyclic order.
*) See, for example, I. M. Yaglom, Geometric Transformations II, Moscow, Gostekhizdat, 1956, §§ 1 and 4 of Chapter II; H. S. M. Coxeter, Introduction to Geometry, Chapter 6; G. Rademacher, O. Toeplitz, Numbers and Figures, Topic 18; D. Hilbert, S. Cohn-Vossen, Anschauliche Geometrie, Moscow, Gostekhizdat, 1951, § 36.
Fig. 41.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
32. Geodesics. This problem does not require knowledge of mathematics. Let's place a rubber band (so-called "prescription", used in pharmacies for packaging medicines) on a stationary cube in such a way that it holds on the cube and does not cross itself.
The line along which this rubber band will lie is called a geodesic line.
1) How many times will all geodesic lines cover the surface of the cube (i.e., how many geodesic lines will pass through each point on the surface of the cube)?
2) How many different families of geodesic lines cover the surface of the cube?
|
32. We will prove that through each point on the surface of a cube, there pass four different geodesics, and in total, we have seven families of geodesic lines.
If we assume that the cube is smooth, then a rubber band wrapped around it will be arranged in such a way that the perimeter of the polygon it forms will reach a minimum.
Fig. 42.
Three types of such positions, and therefore, three families of such geodesics, are shown in Fig. 43; they lie in planes parallel to the faces of the cube. To convince ourselves that there are other families of geodesics, let's cut the cube with a plane parallel to the diagonal of the base (Fig. 44),
Then, using the notations indicated in Fig. 44, we will have:
$$
\begin{gathered}
x+y=a \\
P K=a \sqrt{2}-2 x \operatorname{tg} \alpha, \quad K L=x \sqrt{1+2 \operatorname{tg}^{2} \alpha} \\
M N=a \sqrt{2}-2 y \operatorname{tg} \alpha, \quad L M=y \sqrt{1+2 \operatorname{tg}^{2} \alpha}
\end{gathered}
$$
and for the perimeter \( p \) of the hexagon \( K L M N O P \), we get:
$$
p=2 a \sqrt{2}-2 a \operatorname{tg} \alpha+2 a \sqrt{1+2 \operatorname{tg}^{2} \alpha}
$$
Thus, the perimeter \( p \) depends only on the angle \( \alpha \) and remains the same in all parallel planes; this perimeter
Fig. 43.
Fig. 44.
reaches a minimum if \( \operatorname{tg} \alpha=1 / \sqrt{2} \). The sides of the hexagon \( K L M N O P \) will be parallel to the diagonals of the faces of the cube, and this hexagon represents a geodesic. There are four families of such geodesics, as indicated in Fig. 45, and therefore, together with the previous three, there are a total of seven families.
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
34. Unfolding a Cube. Models of polyhedra are made from flat nets. In a net, faces are adjacent to each other along edges, and the model is constructed by folding the cardboard net along the edges. A regular tetrahedron has two such different nets. How many does a cube have?
|
34. All existing nets (a total of 11) are shown in Fig. 49. The first six solutions give nets in which four faces of the cube are arranged in one strip of the net. No other solutions of this type exist. The next four nets are those in which there are three faces in one
Fig. 48.
Fig. 49.
strip, but not four faces. Finally, in the last net, there are no three faces in any strip.
|
11
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
87. Four dogs. Four dogs $A, B, C$, and $D$ stand at the corners of a square meadow and suddenly start chasing each other as indicated by the arrows in Fig. 2.
Each dog runs straight towards the next: $A$ towards $B$, $B$ towards $C$, $C$ towards $D$, and $D$ towards $A$. The side of the meadow is 100 m, and the speed of the dogs is $10 \mathrm{m} / \mathrm{s}$.
After what time interval will the dogs meet? Will their paths intersect and where? What is the length of each path?
Fig. 2.
|
87. Since each dog runs at a right angle to the direction of the dog chasing it, and the one chasing runs straight toward the fleeing one, the chasing dog approaches the next dog at a speed of $10 \mathrm{m} /$ s and will catch it after 10 seconds. As a result, the path of each dog is $100 \mathrm{~m}$. At every moment, the four dogs form a square. This square rotates and shrinks: its sides decrease uniformly at a speed of $10 \mathrm{m} /$. . The paths will intersect at the center $S$ of the initial square. These will be curved lines (logarithmic spirals). They will not intersect earlier, because if any dog had crossed another's path, it would mean that it had been in that place earlier, which is impossible due to the equal distance of all dogs from $S$ at every moment.
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
88. The Chase (I). Ship $P$ has spotted ship $Q$, which is sailing in a direction perpendicular to $P Q$, maintaining its course. Ship $P$ is chasing $Q$, always heading directly towards $Q$; the speed of both ships is the same at any moment (but can vary over time). Without calculations, it is clear that $P$ is sailing along a curved path; if the chase lasts long enough, the trajectory of the pursuing ship and the trajectory of the fleeing ship will eventually become almost identical. What will then be the distance $P Q$, if initially it was 10 nautical miles?
|
88. Let $\alpha$ denote the instantaneous angle between the direction $P Q$ and the path of ship $Q$ (Fig. 163), and $v$ - the speed of ships $P$ and $Q$ at that moment. The mutual approach of the ships is influenced by the speed $v$ of ship $P$, directed towards $Q$, and the component $v \cos \alpha$ of the speed of ship $Q$, both in the same direction. Therefore, the speed of approach of the ships is $v(1-\cos \alpha)$.
The projection $S$ of point $P$ on the path of ship $Q$ moves along this path with a speed of $v \cos \alpha$, while ship $Q$ moves away with a speed of $v$, so the distance $S Q$ increases at a speed of $v(1-\cos \alpha)$. Since the distance $P Q$ decreases, as we noted earlier, at the same speed, the sum $P Q + S Q$ is constant, and therefore equals 10 miles, as it did at the initial moment.
After an infinitely long
Fig. 163. time, $P$ will coincide with $S$ and will be equal to $P Q + S Q = 2 P Q = 10$ miles, and $P Q = 5$ miles.
|
5
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 2. Let's determine in which numeral system the following multiplication was performed: $352 \cdot 31=20152$.
|
Solution. Let $x$ be the base of the numeral system, then the given equality can be written in the form of an equation
$$
\left(3 x^{2}+5 x+2\right)(3 x+1)=2 x^{4}+x^{2}+5 x+2
$$
By performing the multiplication and combining like terms, we get:
$$
2 x^{4}-9 x^{3}-17 x^{2}-6 x=0
$$
It is clear that $x \neq 0$ and therefore the equation will take the form:
$$
2 x^{3}-9 x^{2}-17 x-6=0
$$
Since $x$ as the base of the numeral system can only take natural values, among the divisors of the constant term, only the numbers $1,2,3,6$ need to be tested. If we also consider that the digit 5 is the largest in the given equality, then only $x=6$ needs to be checked.
Let's divide the polynomial $2 x^{3}-9 x^{2}-17 x-6$ by $(x-6)$ using Horner's scheme:
6 | 2 | -9 | -17 | -6 |
| ---: | ---: | ---: | ---: |
| 2 | 3 | 1 | 0 |
We obtained zero as the remainder. This means that $x=6$ is a root of the obtained equation and is the base of the desired numeral system.
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 2. Let's find the sum of the cubes of the roots of the equation
$$
x^{3}+2 x^{2}+x-3=0
$$
|
S o l u t i o n. This can be done in various ways, for example, by sequentially calculating the sum of the squares of the roots and then the sum of the cubes of the roots. However, we will use a frequently applied identity in mathematics:
$$
\begin{aligned}
a^{3}+b^{3}+c^{3}-3 a b c= & (a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-\right. \\
& -a c-b c) .
\end{aligned}
$$
In our case, this identity will take the form:
$$
\begin{gathered}
x_{1}^{3}+x_{2}^{3}+x_{3}^{3}=3 x_{1} x_{2} x_{3}+\left(x_{1}+x_{2}+x_{3}\right)\left(x_{1}^{2}+x_{2}^{2}+\right. \\
\left.+x_{3}^{2}-x_{1} x_{2}-x_{2} x_{3}-x_{3} x_{1}\right)
\end{gathered}
$$
We will express \( S_{3}=x_{1}^{3}+x_{2}^{3}+x_{3}^{3} \) in terms of \( \sigma_{1}, \sigma_{2} \) and \( \sigma_{3} \), where \( \sigma_{1}=x_{1}+x_{2}+x_{3}, \sigma_{2}=x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}, \sigma_{3}=x_{1} x_{2} x_{3} \).
The sum \( x_{1}^{2}+x_{2}^{2}+x_{3}^{2} \) can be easily expressed in terms of \( \sigma_{1} \) and \( \sigma_{2} \):
$$
x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=\sigma_{1}^{2}-2 \sigma_{2}
$$
Finally, we get:
$$
x_{1}^{3}+x_{2}^{3}+x_{3}^{3}=\sigma_{1}^{3}-3 \sigma_{1} \sigma_{2}+3 \sigma_{3} .
$$
Since by the condition (by Vieta's formulas) \( \sigma_{1}=-2 \),
$$
\sigma_{2}=1, \sigma_{3}=3, \text { then } x_{1}^{3}+x_{2}^{3}+x_{3}^{3}=7
$$
E x a m p l e. Solve the system of equations:
$$
\left\{\begin{array}{l}
x+y+z=2 \\
x^{2}+y^{2}+z^{2}=6 \\
x^{3}+y^{3}+z^{3}=8
\end{array}\right.
$$
S o l u t i o n. Consider the triplet of variables \( x, y, z \) as the roots of some cubic equation \( t^{3}+a t^{2}+b t+c=0 \) and find the coefficients \( a, b, c \).
By Vieta's formulas
$$
\sigma_{1}=-a, \quad \sigma_{2}=b, \quad \sigma_{3}=-c
$$
But from the previous example,
$$
\begin{aligned}
x^{2}+y^{2}+z^{2}=\sigma_{1}^{2}-2 \sigma_{2} & =a^{2}-2 b \\
x^{3}+y^{3}+z^{3}=\sigma_{1}^{3}-3 \sigma_{1} \sigma_{2}+3 \sigma_{3} & =-a^{3}+3 a b-3 c
\end{aligned}
$$
Substituting these expressions into the original system of equations, we get:
$$
\left\{\begin{array}{l}
-a=2 \\
a^{2}-2 b=6 \\
-a^{3}+3 a b-3 c=8
\end{array}\right.
$$
from which we easily find: \( a=-2, b=-1, c=2 \).
Therefore, the desired equation will be:
$$
t^{3}-2 t^{2}-t+2=0
$$
or
$$
\left(t^{2}-1\right)(t-2)=0
$$
Its roots are: \( -1, 1, 2 \). Since the original system is symmetric, any combination of this triplet of numbers is its solution.
## EXERCISES
11. Formulate an equation whose roots are the lengths of the radii \( r_{a}, r_{b}, r_{c} \) of the excircles.
12*. Formulate a cubic equation whose roots are
$$
\cos \frac{\pi}{7}, \cos \frac{3 \pi}{7}, \cos \frac{5 \pi}{7}
$$
13*. Find the sum of the 16th powers of the roots of the equation
$$
x^{3}-x+1=0 \text {. }
$$
14. Calculate the area of a quadrilateral whose side lengths are the roots of the equation \( x^{4}+k x^{3}+m x^{2}+n x+5=0 \), and the sum of the measures of the opposite angles is \( \pi \).
15. If the equation \( a x^{3}-3 b x^{2}+3 c x-d=0 \) has roots \( \frac{b}{a} \), \( \frac{c}{b} \), \( \frac{d}{c} \), then these roots are either equal or form an arithmetic progression. Prove it.
16. Calculate the values of the expressions: \( x_{1}^{2} x_{2}^{2}+x_{2}^{2} x_{3}^{2}+x_{3}^{2} x_{1}^{2} \); \( x_{1}^{2} x_{2}+x_{1} x_{2}^{2}+x_{2}^{2} x_{3}+x_{2} x_{3}^{2}+x_{1}^{2} x_{3}+x_{1} x_{3}^{2} \), if \( x_{1}, x_{2}, x_{3} \) are the roots of the equation \( x^{3}-x+3=0 \).
17. Find the real roots of the systems of equations:
a) \( \left\{\begin{aligned} x+y+z & =4 \\ x y+y z+z x & =5 \\ \frac{1}{x}+\frac{1}{y}+\frac{1}{z} & =\frac{5}{2}\end{aligned}\right. \)
b) \( \left\{\begin{aligned} x^{3}+y^{3}+z^{3} & =\frac{73}{8}, \\ x y+x z+y z & =x+y+z \\ x y z & =1 .\end{aligned}\right. \)
18. Given the equation
$$
x^{3}+p x^{2}+q x+z=0
$$
Find the relationship between the coefficients under which one root is equal to the sum of the other two.
19. Given the equation \( x^{4}+a x^{3}+b x^{2}+c x+d=0 \), in which: a) \( x_{1}+x_{2}=x_{3}+x_{4} \); b) \( x_{1} x_{2}=x_{3} x_{4} \). Find the relationship between the coefficients. Find ways to solve the equation under these conditions.
20. Determine the condition that the coefficients of the equation \( x^{3}+p x^{2}+q x+r=0 \) must satisfy for its three roots to form an arithmetic progression.
21. What necessary and sufficient condition must the coefficients of the equation \( x^{8}+p x^{2}+q x+r=0 \), where \( r \neq 0 \), \( q \neq 0 \), satisfy if its roots are related by the equation \( \frac{1}{x_{1}}+\frac{1}{x_{2}}=\frac{1}{x_{3}} \) ?
22. Determine \( \lambda \) so that one of the roots of the equation \( x^{3}+7 x+\lambda=0 \) is twice the second.
12
Problems on forming equations of the 3rd and 4th degrees. The solutions to the geometric problems presented in this section mainly reduce to solving reciprocal or other equations, which in turn can be appropriately reduced to quadratic equations.
Let's consider this on a specific example.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 1. Let's find pairs of real numbers $x$ and $y$ that satisfy the equation
$$
x^{2}+4 x \cos x y+4=0
$$
|
S o l u t i o n. Several methods can be proposed to solve the given equation.
1st method. Since the equation is quadratic (if we do not consider $x$ under the cosine sign), we can express $x$ in terms of trigonometric functions of the angle $x y$:
$$
x=-2 \cos x y \pm \sqrt{4 \cos ^{2} x y-4}=-2 \cos x y \pm 2 \sqrt{-\sin ^{2} x y}
$$
Since by condition $x$ and $y$ are real numbers, then $-\sin ^{2} x y \geqslant 0$. This inequality is only satisfied when $\sin x y=0$, i.e., when $x y=k \pi, k \in \mathbf{Z}$.
Now we have:
$$
x=-2 \cos x y=-2 \cos k \pi .
$$
If $k$ is even, then $\cos k \pi=1$ and $x=-2$, so $y=n \pi, n \in \mathbf{Z}$; if $k$ is odd, then $\cos k \pi=-1$ and $x=2$, so $y=\frac{\pi}{2}(2 m+$ $+1)$, where $m \in \mathbf{Z}$.
2nd method. Transform the equation so that the left side becomes a sum of non-negative numbers, for example by adding and correspondingly subtracting $4 \cos ^{2} x y$. We get:
$$
(x+2 \cos x y)^{2}+4\left(1-\cos ^{2} x y\right)=0
$$
or
$$
(x+2 \cos x y)^{2}+4 \sin ^{2} x y=0
$$
Since the sum of squares of real numbers equals zero only when each term is zero, we arrive at the system:
$$
\left\{\begin{array}{c}
x+2 \cos x y=0 \\
\sin x y=0
\end{array}\right.
$$
Solving this system, we find the same values of $x$ and $y$.
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 5. Let's find all values of $a$ for which the system
$$
\left\{\begin{array}{l}
2^{b x}+(a+1) b y^{2}=a^{2} \\
(a-1) x^{3}+y^{3}=1
\end{array}\right.
$$
has at least one solution for any value of $b,(a, b, x, y \in \mathbf{R})$.
|
S o l u t i o n. Suppose there exists some value $a$ for which the system has at least one solution, for example, for $b=0$.
In this case, the system will take the form:
$$
\left\{\begin{array}{l}
1=a^{2} \\
(a-1) x^{3}+y^{3}=1
\end{array}\right.
$$
It is clear that this system is consistent only if $a=1$ or $a=-1$. Therefore, if $b=0$, then for these values of $a$ the original system is consistent. We will now prove that the original system will be consistent for any value of $b$.
32
Let $a=1$. Then the original system will take the form:
$$
\left\{\begin{array}{l}
2^{b x}+2 b y^{2}=1 \\
y^{3}=1
\end{array}\right.
$$
From the second equation, we find: $y=1$ (by the condition $y \in \mathbf{R}$). Substituting $y=1$ into the first equation, we get:
$$
2^{b x}=1-2 b
$$
It is easy to see that this equation has no solutions for $b \geqslant \frac{1}{2}$. Therefore, when $a=1$, the system has solutions, but not for any value of $b$, i.e., the value $a=1$ does not satisfy the condition of the problem.
Now let's check $a=-1$. Then the original system will take the form:
$$
\left\{\begin{array}{l}
2^{b x}=1 \\
-2 x^{3}+y^{3}=1
\end{array}\right.
$$
It is completely obvious that the first equation has a solution $x=0$ for any $b$. Then $y=1$. Therefore, the condition of the problem is satisfied only by $a=-1$.
## EXERCISES
57. Prove that the equation
$$
x^{2}-x \sin x y+1=0
$$
has no solutions.
58. Find all pairs of real values of $x$ and $y$ that satisfy the equations:
a) $\sin 2 x+\cos (x+y)=2$
b) $4^{\sin x}-2^{1+\sin x} \cos x y+2^{|y|}=0$;
c) $\operatorname{tg}^{2} x+2 \operatorname{tg} x(\sin x+\cos y)+2=0$;
d) $\left(\sin ^{2} x+\frac{1}{\sin ^{2} x}\right)^{2}+\left(\cos ^{2} x+\frac{1}{\cos ^{2} y}\right)^{2}=12+\frac{1}{2} \sin x y$;
e) $\operatorname{tg}^{4} x+\operatorname{tg}^{4} y+2 \operatorname{ctg}^{2} x \operatorname{ctg}^{2} y=3+\sin ^{2}(x+y)$.
59. Solve the inequalities:
a) $\cos x \geqslant y^{2}+\sqrt{y-x^{2}-1}$
b) $y \geqslant\left|\sec ^{2} x\right|+\sqrt{1-y-x^{2}}$
c) $x-|y| \geqslant 1+\sqrt{x^{2}+y^{2}-1}$.
60. Find the general form of the solution of the system:
$$
\left\{\begin{array}{l}
\operatorname{tg} x=\operatorname{tg}(y-z), \\
\operatorname{tg} y=\operatorname{tg}(z-x) .
\end{array}\right.
$$
61. Find the real solutions of the systems:
a) $\left\{\begin{array}{l}x^{2}+4 y^{2}+5=4 z \\ x-y \geqslant z\end{array}\right.$
2 Order 454
b) $\left\{\begin{array}{l}x^{2}+y^{2}+20=z \\ 8 x+4 y \geqslant z .\end{array}\right.$
62. Find the real solutions of the system:
$$
\left\{\begin{array}{l}
x+y+z=2 \\
2 x y-z^{2}=4
\end{array}\right.
$$
63. Prove that if three numbers $x, y, z$ satisfy the system of equations:
$$
\left\{\begin{array}{l}
x+y+z=a \\
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{a}
\end{array}\right.
$$
then at least one of these numbers is equal to $a$.
64. Find all values of $a$ for which the system
$$
\left\{\begin{array}{l}
2^{|x|}+|x|=y+x^{2}+a \\
x^{2}+y^{2}=1
\end{array}\right.
$$
has only one solution.
65. Find all values of $a$ and $b$ for which the system
$$
\left\{\begin{array}{l}
\left|\frac{x^{y}-1}{x^{y}+1}\right|=a \\
x^{2}+y^{2}=b
\end{array}\right.
$$
has only one solution ( $a, b, x, y \in \mathbf{R}, x>0, x \neq 1$ ).
66. Find all values of $a$ for which the system
$$
\left\{\begin{array}{l}
\left(x^{2}+1\right)^{a}+\left(b^{2}+1\right)^{y}=2 \\
a+b x y+x^{2} y=1
\end{array}\right.
$$
has at least one solution for any $b(a, b, x, y \in \mathbf{R})$.
67. Determine for which values of the parameter $a$ the system
$$
\left\{\begin{array}{l}
a x^{2}+a-1=y-|\sin x| \\
\operatorname{tg}^{2} x+y^{2}=1
\end{array}\right.
$$
has a unique real solution.
68. Given the system:
$$
\left\{\begin{array}{l}
x^{y}=a, x>0 \\
\operatorname{arctg} x=\frac{\pi}{4}+y
\end{array}\right.
$$
For which values of $a$ does the system have a unique solution?
69. Find all values of $a$ for which the system
$$
\left\{\begin{array}{l}
x^{3}-a y^{3}=\frac{1}{2}(a+1)^{2} \\
x^{3}+a x^{2} y+x y^{2}=1
\end{array}\right.
$$
has at least one solution and any solution satisfies the equation $x+y=0 \quad(a, x, y \in \mathbf{R})$.
34
70. Given the system of equations:
$$
\left\{\begin{array}{l}
\frac{x}{y}+\sin x=a \\
\frac{y}{x}+\sin y=a
\end{array}\right.
$$
For which values of $a$ does the system have a unique solution satisfying the conditions: $0 \leqslant x \leqslant 2 \pi, 0 \leqslant y \leqslant 2 \pi$ ?
## Chapter II
## POWER MEAN AND ITS APPLICATIONS
The expression of the form
$$
C_{\alpha}(a)=\left(\frac{a_{1}^{\alpha}+a_{2}^{\alpha}+\ldots+a_{n}^{\alpha}}{n}\right)^{\frac{1}{\alpha}}
$$
is called the power mean of order $\alpha$ of positive numbers $a_{1}, a_{2}, \ldots, a_{n}$.
In particular, for $\alpha=-1$ we get the harmonic mean
$$
C_{-1}(a)=\frac{n}{\frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{n}}}
$$
for $\alpha=1$ we get the arithmetic mean
$$
C_{1}(a)=\frac{a_{1}+a_{2}+\ldots+a_{n}}{n}
$$
for $\alpha=2$ we get the quadratic mean
$$
C_{2}(a)=\sqrt{\frac{a_{1}^{2}+a_{2}^{2}+\ldots+a_{n}^{2}}{n}}
$$
Let us note some properties of the power mean:
$1^{0}$. The following equality holds:
$$
\lim _{\alpha \rightarrow 0} C_{\alpha}(a)=G
$$
where $G=\sqrt[n]{a_{1} a_{2} \ldots a_{n}}$ is the geometric mean. Therefore, we adopt
$$
C_{0}(a)=\sqrt[n]{a_{1} a_{2} \ldots a_{n}}
$$
$2^{0}$. For any real $\alpha$ and $\beta$, such that $\alpha \leqslant \beta$, the following inequality holds (monotonicity property):
$$
C_{\alpha}(a) \leqslant C_{\beta}(a)
$$
$3^{0}$. Let $\min a$ be the smallest among the numbers $a_{1}, a_{2}, \ldots a_{n}$, and $\max a$ be the largest among these same numbers. Then
$$
\lim _{\alpha \rightarrow-\infty} C_{\alpha}(a)=\min a, \lim _{\alpha \rightarrow+\infty} C_{\alpha}(a)=\max a
$$
2*
From this, the following chain of inequalities follows:
$\min a \leqslant \ldots \leqslant C_{-2} \leqslant C_{-1} \leqslant C_{0} \leqslant C_{1} \leqslant C_{2} \leqslant \ldots \leqslant \max a$.
Equality occurs only when $a_{1}=a_{2}=\ldots=a_{n}$.
## § 1. Proof of Inequalities
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 3. Compare the number $a$ with one, if
$$
a=0.99999^{1.00001} \cdot 1.00001^{0.99999} .
$$
|
S o l u t i o n. Let's represent the numbers in the given expression as follows:
$$
0.99999=1-\alpha, \quad 1.00001=1+\alpha
$$
where $\alpha=0.00001$. Then
$$
a=(1-\alpha)^{1+\alpha}(1+\alpha)^{1-\alpha}=\left(1-\alpha^{2}\right)\left(\frac{1-\alpha}{1+\alpha}\right)^{\alpha}
$$
Since $1-\alpha^{2}<1$, then $a<1$
b) $\frac{1}{2}<5$
$$
156. Prove that
$$
\begin{gathered}
\log _{\sqrt{2}+\sqrt{3}}(4 \sqrt{2}+3 \sqrt{\overline{3}}) \cdot \log _{\sqrt{6}+1}(\sqrt{3}-\sqrt{2})+ \\
+\log ^{2} \sqrt{6}+7 \\
(2 \sqrt{6}+5)+1=0
\end{gathered}
$$
## § 2. Systematic Fractions and Irrationalities
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
E x a m p l e 1. Let's find the limit of the sum
$$
S_{n}=\frac{3}{4}+\frac{5}{36}+\ldots+\frac{2 n+1}{n^{2}(n+1)^{2}}
$$
as $n \rightarrow \infty$.
|
S o l u t i o n. First, let's try to simplify $S_{n}$. Since the fraction 56
$\frac{2 k+1}{k^{2}(k+1)^{2}}$ can be represented as the difference $\frac{1}{k^{2}}-\frac{1}{(k+1)^{2}}$, then
$$
\begin{gathered}
\frac{3}{4}=1-\frac{1}{2^{2}} \\
\frac{5}{36}=\frac{1}{2^{2}}-\frac{1}{3^{2}} \\
\frac{2 n+1}{n^{2}(n+1)^{2}}=\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}}
\end{gathered}
$$
By adding these equalities, we get:
$$
S_{n}=1-\frac{1}{(n+1)^{2}}
$$
Then
$$
\lim _{n \rightarrow \infty} S_{n}=\lim _{n \rightarrow \infty}\left(1-\frac{1}{(n+1)^{2}}\right)=1
$$
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 2. Let $y \neq-1$. We set,
$$
x_{1}=\frac{y-1}{y+1}, \quad x_{2}=\frac{x_{1}-1}{x_{1}+1}, \quad x_{3}=\frac{x_{2}-1}{x_{2}+1}, \ldots
$$
What is $y$ if $x_{1978}=-\frac{1}{3}$?
|
Solution. Substituting the value of $x_{1}$ into the second equality, after simplifications we get:
$$
x_{2}=-\frac{1}{y}
$$
Further,
$$
x_{3}=\frac{y+1}{1-y}, \quad x_{4}=y, \quad x_{5}=\frac{y-1}{y+1}=x_{1}
$$
Therefore,
$$
x_{5}=x_{1}, \quad x_{6}=x_{2}, \quad x_{7}=x_{3}, \quad x_{8}=x_{4}, \ldots, \quad x_{1978}=x_{2}=-\frac{1}{y}
$$
from which $y=3$.
## EXERCISES
174. Let a number $N$ be given and a sequence of positive numbers $x_{1}, x_{2}, \ldots, x_{n}$, where each subsequent term is formed by the rule
$$
x_{n}=\frac{1}{2}\left(x_{n-1}+\frac{N}{x_{n-1}}\right)
$$
Prove that
$$
\lim _{n \rightarrow \infty} x_{n}=\sqrt{N}
$$
175. A sequence is defined by the recurrence formula
$$
u_{n}=(\alpha+\beta) u_{n-1}-\alpha \beta u_{n-2}
$$
and initial values $u_{1}=\alpha+\beta, u_{2}=\frac{\alpha^{3}-\beta^{3}}{\alpha-\beta}$. Find the general term of the sequence.
176. Two sequences are given: $x_{0}, x_{1}, \ldots, x_{n}, \ldots\left(x_{0}>0\right)$ and $y_{0}, y_{1}, \ldots, y_{n}, \ldots\left(y_{0}>0\right)$, where
$$
x_{n}=\frac{x_{n-1}+y_{n-1}}{2}, \quad y_{n}=\sqrt{x_{n-1} y_{n-1}}
$$
Prove that the limits $\lim _{n \rightarrow \infty} x_{n}$ and $\lim _{n \rightarrow \infty} y_{n}$ exist and are equal to each other.
177. Two positive numbers $a$ and $b(a>b)$ and two sequences are given:
$$
\begin{array}{ll}
a_{1}=\frac{a+b}{2}, & b_{1}=\frac{2 a b}{a+b} \\
a_{2}=\frac{a_{1}+b_{1}}{2}, & b_{2}=\frac{2 a_{1} b_{1}}{a_{1}+b_{1}} \\
\cdots \cdot \cdot \cdot \cdot, & . \cdot \cdot \cdot \cdot \\
a_{n+1}=\frac{a_{n}+b_{n}}{2}, & b_{n+1}=\frac{2 a_{n} b_{n}}{a_{n}+b_{n}}
\end{array}
$$
Find the limits of these sequences.
178. Given 1979 numbers: $x_{1}=0, x_{2}, \ldots, x_{1978}, x_{1979}=0$. Moreover, $x_{i}=\frac{x_{i-1}+x_{i+1}}{4}+1$ for any $i=2,3, \ldots, 1978$. Prove that:
a) $0 \leqslant x_{i} \leqslant 2$ for $i=1,2, \ldots, 1979$
b) $x_{1}x_{992}>\ldots>x_{1979}$
c) $x_{1}=x_{1979}, x_{2}=x_{1978}, \ldots$.
179. Find the sum $\frac{1}{2!}+\frac{2}{3!}+\ldots+\frac{n}{(n+1)!}$ and its limit as $n \rightarrow \infty$.
180. Compute the limits:
a) $\lim \sqrt[n]{n}$;
b) $\lim _{n \rightarrow \infty} \sqrt[n]{\frac{1}{n!}}$
c) $\lim _{n \rightarrow \infty}\left(\frac{5}{9} \cdot \frac{14}{20} \cdot \frac{27}{35} \ldots \frac{2 n^{2}-n-1}{2 n^{2}+n-1}\right)$.
181. Prove that for $x \neq 0$ we have:
$$
\lim _{n \rightarrow \infty}\left(\cos \frac{x}{2} \cdot \cos \frac{x}{4} \ldots \cos \frac{x}{2^{n}}\right)=\frac{\sin x}{x}
$$
182. Find $\lim _{n \rightarrow \infty} x_{n}$, if:
58
a) $x_{n}=\sqrt{\underbrace{2+\sqrt{2+\sqrt{2+\ldots+\sqrt{2}}}}_{n \text { times }}} ;$
b) $x_{n}=\sqrt{\underbrace{2 \sqrt{2 \sqrt{2 \ldots \sqrt{2}}}}_{n \text { times }}}$;
## § 4. Various Algebraic Problems
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Example 2. Let's calculate the sum
$$
a^{2000}+\frac{1}{a^{2000}}
$$
if $a^{2}-a+1=0$.
|
S o l u t i o n. From the equality $a^{2}-a+1=0$, it follows that:
$$
a-1+\frac{1}{a}=0 \text { or } a+\frac{1}{a}=1
$$
Moreover,
$$
a^{3}+1=(a+1)\left(a^{2}-a+1\right)=0
$$
from which $a^{3}=-1$. Now we have:
$$
\begin{aligned}
& a^{2000}+\frac{1}{a^{2000}}=\left(a^{3}\right)^{666} a^{2}+\frac{1}{\left(a^{3}\right)^{666} a^{2}}=a^{2}+\frac{1}{a^{2}}=\left(a+\frac{1}{a}\right)^{2}-2= \\
& =1-2=-1
\end{aligned}
$$
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Points on a Line. If 10 points are placed at equal intervals on a line, they will occupy a segment of length s, and if 100 points are placed, the segment will have a length S. How many times greater is S than s?
|
6. Between ten points there are nine intervals, and between a hundred points - ninety-nine. Therefore, $S$ is greater than $s$ by 11 times.
|
11
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Stereometric problem. How many faces does a hexagonal pencil have?
|
7. It is important to ask: which pencil? If the pencil has not been sharpened yet, then 8, otherwise there may be variations...
Translating the text as requested, while preserving the original line breaks and formatting.
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
12. The Musketeers' Journey. The distance between Athos and Aramis, riding on the road, is 20 leagues. In one hour, Athos travels 4 leagues, and Aramis - 5 leagues. What distance will be between them after an hour?
|
12. Well, in which direction was each of the musketeers traveling? The problem statement does not mention this. If they were traveling towards each other, the distance between them would be 11 leagues. In other cases (make a diagram!), the possible answers are: 29 leagues; 19 leagues; 21 leagues.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
13. An Arabic Tale. A Flock of Pigeons
$>\Delta \theta \Pi \mathrm{V} \oplus \theta\mathbf{v} \Pi \square \nabla \square \Lambda$
$\oplus \Lambda \nabla \theta \Pi \oplus \mathbf{V V} \varnothing \odot$
ロจ৫ఠ<>VIVOO flew up to a tall tree.
Some of the pigeons perched on the branches, while others settled under the tree. "Those perched on the branches said to those below: If one of you flew up to us, then you would be three times fewer than all of us together, and if one of us flew down to you, then we would be equal in number." How many pigeons were perched on the branches and how many were under the tree? (Tales from One Thousand and One Nights. Night 458)
|
13. From the condition of the problem, it is clear that the number of pigeons sitting on the branches is two more than those sitting below. Further, it follows from the condition that after one of the pigeons flew up to the branch, the number of pigeons sitting on the branch became twice as many as those sitting on the ground. In addition, at this point, there were four more of them. Therefore, 7 pigeons are sitting on the tree, and 5 under the tree.
|
7
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In the government of the country of knights and liars, there are 12 ministers. Some of them are liars, and the rest are knights. Once, at a government meeting, the following opinions were expressed: the first minister said, “There is not a single honest person here,” the second said, “There is no more than one honest person here,” the third said, “There are no more than two honest people here,” and so on until the twelfth, who said, “There are no more than eleven honest people here.” How many liars are in the government of the country?
23 In the country of knights and liars
|
4. Note that the number of true statements must match the number of honest people in the government. Further, if a statement from any minister is true, then the statements of each minister who spoke after him are also true. In this case, the only statement that would not lead to a contradiction is: "there are no more than 6 honest people," since in this case, exactly 6 statements would be true. Therefore, there are exactly 6 liars in the government.
|
6
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Around a round table, eight people are sitting, each of whom is either a knight or a liar. When asked who their neighbors are, each of them answered: “My neighbors are a liar and a knight.” How many of them were liars? How would the answer change if nine people were sitting at the table?
|
5. 6. At the table, there is at least one liar. Indeed, if only knights were sitting at the table, each knight's statement "next to me sits a knight and a liar" would be false, which is impossible. 2. The neighbors
Fig. 1
Who can sit here? Both possible assumptions lead to a contradiction
Fig. 2
of a liar can be either two liars or two knights. 3. If both neighbors of a liar are liars, then further around the table sit only liars, otherwise, one of the liars' statements "next to me sits a knight and a liar" would be true, which is impossible. Thus, one of the possible answers is all liars. 4. If the neighbors of a liar are knights, then after each knight must sit another knight, then a liar, then again two knights, then a liar, and so on. If there are 9 people at the table, then there are 3 liars (Fig. 1), if 8 people, then we get a contradiction (Fig. 2$)^{1}$.
Answer: 1. If there are 8 people at the table, then all are liars. 2. If there are 9 people at the table, then the number of liars is either 9 or 3.
|
6
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
14. Insert parentheses in the equation: $1: 2: 3: 4: 5: 6: 7: 8: 9: 10=7$, to make it true.
|
14. Answer: $1: 2: 3: 4: 5:(6: 7: 8: 9: 10)=7$.
|
7
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. What can you buy for a ruble? Nine boxes of matches cost 9 rubles and some kopecks, while ten such boxes cost 11 rubles and some kopecks. How much does one box cost?
|
6. Note that a box of matches costs more than $\frac{11}{10}$ rubles, but less than $\frac{10}{9}$ rubles. That is, more than 1.10 rubles, but less than 1.111 rubles. Answer: 1 ruble 11 kopecks.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
12. The Big Laundry. After seven hours of washing, the length, width, and height of the soap piece were halved. For how many washes will the remaining soap last?
|
12. Since the length, width, and height of the soap piece have been halved, its volume has decreased by 8 times, meaning that in 7 hours, the soap piece has decreased by $\frac{7}{8}$ of its volume (by $\frac{1}{8}$ of its volume per hour). Therefore, the soap will last for one more hour of heavy washing.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. $о$ about the fisherman and the fish. When asked how big the fish he caught was, the fisherman said: “I think that its tail is 1 kg, the head is as much as the tail and half of the body, and the body is as much as the head and the tail together.” How big is the fish?
|
3. Let $2 x$ kg be the weight of the torso, then the head will weigh $x+1$ kg. From the condition that the torso weighs as much as the head and tail together, we get the equation: $2 x=x+1+1$. From this, $x=2$, and the whole fish weighs - 8 kg.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11. The Clock Hands Problem. At what time after 12:00 will the minute hand first catch up with the hour hand?
|
11. At 13:00, the minute hand will lag behind the hour hand by 5 minute divisions. Before the "meeting," the hour hand will travel $x$ divisions, and the minute hand will travel $12x$ divisions. From the equation $x + 5 = 12x$, we get that $x = \frac{5}{11}$ minute divisions on the clock face. Answer: 1 hour $5 \frac{5}{11}$ minutes.
|
1
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Divide equally. There is milk in an eight-liter bucket. How can you measure out 4 liters of milk using a five-liter bucket and a three-liter jar?
|
4. It is convenient to write the solution in the form of a table:
| 8-liter can | 8 | 3 | 3 | 6 | 6 | 1 | 1 | 4 |
| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 5-liter can | 0 | 5 | 2 | 2 | 0 | 5 | 4 | 4 |
| 3-liter jar | 0 | 0 | 3 | 0 | 2 | 2 | 3 | 0 |
|
4
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In a family, there are six children. Five of them are respectively 2, 6, 8, 12, and 14 years older than the youngest, and the age of each child is a prime number. How old is the youngest?
|
4. Suppose the child's age does not exceed 35 years. Let's list all the prime numbers: $2,3,5,7,11,13,17,19,23$, 29,31. It is clear that the age of the younger child is an odd number. The numbers 29 and 31 also do not fit. Let's find the age of the younger child. He cannot be 1 year old, because $1+8=9$. His age cannot end in 3, because $3+2=5$. This leaves $5,7,11,17$ and 19. Not 7 years old, because $7+2=9$. Not 17 years old, because $17+8=25$. Not 19 years old: $19+2=21$. And not 11: $11+14=25$. Therefore, the younger one is 5 years old. Answer: $5,7,11,13,17$ and 19 years.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
13. In the senate of the country of knights and liars - there are 100 senators. Each of them is either a knight or a liar. It is known that: 1. At least one of the senators is a knight. 2. Out of any two arbitrarily chosen senators, at least one is a liar. Determine how many knights and how many liars are in the senate.
|
13. Only one of the senators is a knight. The fact that there is at least one knight among the senators follows from the first statement, the existence of a second knight contradicts the second statement.
|
1
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Scientific organization of labor. There are logs of two types: 6 meters and 7 meters long. They need to be sawn into 1-meter logs. Which logs are more profitable to saw?
|
1. To saw 42 one-meter logs from six-meter logs requires 35 cuts, while from seven-meter logs - 36. It can be considered that sawing six-meter logs is more profitable.
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. To the Mechanics and Mathematics Faculty! Several identical books and identical albums were bought. The books cost 10 rubles 56 kopecks. How many books were bought if the price of one book is more than one ruble higher than the price of an album, and 6 more books were bought than albums?
|
5. Since each book is more expensive than a ruble, no more than 10 books were bought. Moreover, it is clear that no fewer than 7 books were bought (since at least one album was bought). The number 1056 is divisible by 8 and not divisible by $7,9,10$. Therefore, 8 books were bought. (MSU, Mechanics and Mathematics, 1968)
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. One or two? Let's take all natural numbers from 1 to 1000000 and for each of them, calculate the sum of its digits. For all the resulting numbers, we will again find the sum of their digits. We will continue this process until all the resulting numbers are single-digit. Among the million resulting numbers, 1 and 2 will appear. Which number will be more frequent: 1 or 2?
|
8. Let's use the statement: if a number when divided by 9 has a remainder of \( d \), then the sum of its digits will have the same remainder. Which numbers from 1 to 1000000 are more: those that have a remainder of 1 when divided by 9, or those that have a remainder of 2? In the range from 1 to 999999, there are an equal number of each. 1000000 when divided by 9 gives a remainder of 1. Therefore, 1 is more.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Two detectives met. Here is their dialogue:
- Do you have two sons?
- Yes, they are young, they don't go to school.
- By the way, the product of their ages equals the number of pigeons near us.
- That's not enough information.
- And I named the older one after you.
- Now I know how old they are.
How old are the sons?
|
7. The ages of the sons could have been equal, hence the product of their ages - a perfect square. Upon receiving information about the product of the ages, the second detective could not answer the question, therefore, the product can be factored in two ways. Answer: 1 and 4 years.
|
1
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. In a chess tournament, 8 players participated. They scored $7, 6, 4, 4, 3, 2, 1, 5$ and 0.5 points respectively. How many points did the players who took the first four places lose in matches against the others?
|
7. From the condition of the problem, it follows that the tournament winner and the participant who took second place did not lose a single point in their matches against the others. Further, the participants who shared third and fourth places could have scored 8 points against the rest, but scored 7 (they played one point in their match against each other). Thus, the players who took the first four places lost 1 point in their matches against the others.
73 Tournaments
|
1
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11. Homework. Fedya was supposed to divide a certain number by 4 and add 15 to it, but Fedya multiplied this number by 4 and subtracted 15, yet he still got the correct answer. What was this number?
|
11. Solving the equation: $0.25 x + 15 = 4 x - 15$, we get the answer: 8.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Linear function. The distance between villages $\mathcal{A}$ and $B$ is 3 km. In village $\mathcal{A}-300$ students, and in village $B-$ 200 students. Where should a school be built to minimize the total distance traveled by students on their way to school?
|
4. It is clear that the school should be built on the segment $A B$, but where exactly? Let the distance from village $A$ to the school be $x$, then the total distance traveled by all schoolchildren on the way
77 What is the best?
to school is $f(x)=300 x+200(a-x)=200 a+100 x$, where $a-$ is the length of segment $A B$. The minimum value of $f(x)$ is achieved at $x=0$. Therefore, the school should be built in village $A$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11. A Trip for Milk. Anton went to a dairy store. He had no money, but he had empty bottles - six one-liter bottles (worth 20 kopecks each) and six half-liter bottles (worth 15 kopecks each). In the store, there was milk sold by the liter for 22 kopecks. What is the maximum amount of milk he could bring home? He had no other containers besides the empty bottles.
|
11. By returning six half-liter bottles and one liter bottle, Anton will receive 1 ruble 10 kopecks, which will be the cost of 5 liters of milk. The 5 liters of milk he buys can be carried home in the remaining liter bottles. Let's ensure that he won't be able to carry more than 5 liters. If he returns not one liter bottle, but more, then to gather the cost of at least 5 liters of milk, he will need to return at least 5 more half-liter bottles, and the capacity of the remaining bottles will not exceed 4.5 liters (this is verified by enumeration).
|
5
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
12. Given a 1998-digit number, every two adjacent digits of which form a two-digit number divisible by 17 or 23. The last digit of the number is 1. What is the first?
|
12. Let's start recording this number from the end. At some point, we will notice that the number has the form ...92346...9234692346851. Further, we see that the digits 92346 are repeating, so we subtract from 1998 the number of digits that do not belong to this cycle - 3. We divide the resulting number by 5 (the number of digits in the cycle). This number will divide evenly. Therefore, the first digit is 9.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
13. The warehouse has nails in boxes weighing $24, 23, 17$ and 16 kg. Can the warehouse keeper issue 100 kg of nails from the warehouse without opening the boxes?
|
13. For example: 4 boxes - at 17 kg each and 2 boxes - at 16 kg each.
|
4
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
30. What digit does the number $3^{100}$ end with?
|
30. The number $3^{100}=81^{25}$, and therefore, ends in 1.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
31. What is the remainder when the number $2^{99}$ is divided by 7?
95 More problems!
|
31. Note that $2^{99}=8^{33}=(7+1)^{33}=(7+1) \ldots(7+1)$. Expand the brackets. The resulting terms will be divisible by 7, except for 1. Thus, $2^{99}$ can be written in the form: $7 x$ +1 . Answer: 1.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
39. Calculate: $2379 \cdot 23782378-2378 \cdot 23792379$.
|
39. Let $a=2378$, then the desired expression is: $(a+1)(10000 a+a)-a(10000(a+1)+(a+1))=$ $=(a+1) \cdot 10001 a-a \cdot 10001 \cdot(a+1)=0$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
41.On the farmyard, geese and piglets were wandering around. A boy counted the number of heads, there were 30, then he counted the total number of legs, there were 84. Can you find out how many geese and how many piglets were on the farmyard?
|
41.If only geese were wandering around the farmyard, there would be 60 legs in total, the "extra" legs, which number 24, belong to the piglets - two for each. Therefore, there were 12 piglets, and 18 geese.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
42.A brick weighs 2 kg and half a brick. How much does the brick weigh?
|
42.It follows from the condition that half a brick weighs 2 kg. Therefore, a whole brick weighs 4 kg.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
47. 6 carps are heavier than 10 perches, but lighter than 5 pikes; 10 carps are heavier than 8 pikes. What is heavier: 2 carps or 3 perches?
|
47. Since 6 carp are heavier than 10 perch, it is clear that 6 carp are even heavier than 9 perch. Therefore, 2 carp are heavier than 3 perch. This means that two of the three conditions in the problem are redundant.
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
50.What is the 1997th digit in the decimal expansion of the fraction $\frac{1}{7}=0.142857 \ldots ?$
|
50.If we divide 1 by 7 using long division, we get that $\frac{1}{7}=0.(142857)$. The remainder of 1997 divided by 6 is 5, Therefore, the digit at the 1997th place is 5.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
51. Little One eats a jar of jam in six minutes, while Karlson is twice as fast. How long will it take them to eat the jam together
|
51. The question of the problem can also be formulated as follows: "How long would it take for three Little Ones to eat the jam?" (According to the condition of the problem, Carlsson can be equated to two Little Ones). It is clear that three Little Ones would finish the jam three times faster than one. Answer: in 2 minutes.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
60.Yura left the house for school 5 minutes later than Lena, but walked at twice her speed. How long after leaving will Yura catch up to Lena?
|
60.Let Lena walk $s$ km in 5 minutes. Then in the next 5 minutes, Yura will walk $2 s$ km, and Lena will walk another $s$ km, that is, a total of $2 s$ km. Therefore, in 5 minutes, Yura will catch up with Lena.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
61.If a cyclist rides at a speed of 10 km/h, he will be 1 hour late. If he rides at a speed of 15 km/h, he will arrive 1 hour early. At what speed should he ride to arrive on time?
|
61.If there were two cyclists, with the first one's speed being $10 \mathrm{km} / \mathrm{h}$ and the second one's speed being $15 \mathrm{km} / \mathrm{h}$. Then, according to the problem, if the first cyclist started 2 hours earlier than the second, they would arrive at the destination simultaneously. In this case, in 2 hours, the first cyclist would cover 20 km, and the second cyclist would be able to catch up in 4 hours - already at the final destination. Therefore, the entire distance is 60 km. The question of the problem can be rephrased as: "At what speed should a cyclist travel to cover the entire distance in 5 hours?". Answer: 12 km/h.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
64.A motorboat travels 90 km downstream in the same time it takes to travel 70 km upstream. What distance can a raft drift in the same time?
|
64.Let in $t$ hours the motorboat covers 90 km downstream and 70 km upstream. Then the speed of the motorboat downstream is $\frac{90}{t}$ km/hour, and the speed upstream is $\frac{70}{t}$ km/hour. From this, the doubled speed of the current will be $\frac{90}{t}-\frac{70}{t}=\frac{20}{t}$ km/hour. The speed of the current (and therefore the rafts) will be $\frac{10}{t}$ km/hour, and in $t$ hours they will cover 10 km.
## 107 More problems!
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
69. Someone has 12 pints of wine and wants to pour out half, but he does not have a 6-pint container. However, he has two containers with capacities of 5 and 8 pints. How can he measure exactly 6 pints of wine?
|
69. The solution is visible from the table:
| steps: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 12 l | 12 | 4 | 4 | 9 | 9 | 1 | 1 | 6 |
| 8 l | 0 | 8 | 3 | 3 | 0 | 8 | 6 | 6 |
| 5 l | 0 | 0 | 5 | 0 | 3 | 3 | 5 | 0 |
|
6
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
76. In the box, there are 100 white, 100 red, 100 blue, and 100 black balls. What is the smallest number of balls that need to be pulled out, without looking into the box, to ensure that among them there are at least 3 balls of the same color?
|
76. Answer: 9 balls. If we randomly draw 8 balls, there might not be three balls of the same color (2 white + 2 red + 2 blue + 2 black). If we add one more ball, then there will definitely be 3 balls of the same color.
|
9
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
78. How many circles of radius 1 are needed to cover a square with a side length of $2$?
|
78. A circle of radius 1 can only cover one vertex of a $2 \times 2$ square, so at least four circles are required. However, a circle of unit radius can completely cover a $1 \times 1$ square, so four circles are sufficient. Answer: 4 circles.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
81. In the box, there are pencils: 7 red and 5 blue. In the dark, pencils are taken. How many pencils need to be taken to ensure that there are at least two red and at least three blue among them?
|
81. In order to definitely take no less than two red pencils, you need to take no less than 7 pencils, and to definitely take no less than 3 blue ones, you need to take no less than 10 pencils. Therefore, to definitely take no less than two red and no less than 3 blue, you need to take no less than 10 pencils.
|
10
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 11. Find the maximum value of the function $y=3 x+4 \sqrt{1-x^{2}}$
$$
|
\text { S o l u t i o n. }
$$
First, let's find the domain of the given function: $D(f)=$ $=\left\{x \mid 1-x^{2} \geqslant 0\right\}=\{x \mid-1 \leqslant x \leqslant 1\} ; \quad E(f)=\{y \mid \quad$ the equation $y=3 x+4 \sqrt{1-x^{2}}$ has a solution on $\left.[-1 ; 1]\right\}=\{y \mid$ the equation $y-3 x=4 \sqrt{1-x^{2}}$ has a solution on $\left.[-1 ; 1]\right\}=\{y \mid$ the equation $25 x^{2}-6 x y+\left(y^{2}-16\right)=0, \quad$ where $y \geqslant 3 x$, has a solution on $[-1 ; 1]\}=$
$$
\begin{gathered}
=\left\{y \mid\left(y \geqslant 3 x \cap y^{2}-25 \leqslant 0\right) \cap-1 \leqslant x \leqslant 1\right\}= \\
=\{y \mid(y \geqslant 3 x \cap-5 \leqslant y \leqslant 5) \cap-1 \leqslant x \leqslant 1\}= \\
=\{y \mid-3 \leqslant y \leqslant 5\}
\end{gathered}
$$
16
Therefore, $\quad y_{\text {max }}=5$ (achieved when $\left.\quad x=\frac{3}{5}\right)$.
Let's explain the solution of the system $\left\{\begin{array}{l}y \geqslant 3 x \\ -5 \leqslant y\end{array}\right.$ $\left\{\begin{array}{l}-5 \leqslant y \leqslant 5 \\ -1 \leqslant x \leqslant 1\end{array}\right.$ graphically (Fig. 19).
3 p r o b l e m 12. A cylindrical rod is being turned. The length of the rod is $l$, the diameter of the base is $D, h$-the thickness of the layer being turned. Find the maximum displacement of the center of gravity during the turning of the cylinder ${ }^{1}$.
Fig. 19
$$
\text { S o l u t i o n. }
$$
Introduce a rectangular coordinate system as shown in Fig. 20; $x$ - the length of the turned layer at some moment in time $t$ :
$$
0 \leqslant x \leqslant l
$$
Before turning, the center of gravity of the cylinder was at point $C\left(\frac{l}{2} ; 0\right)$. Then, as the turning progresses, point $C$ will start moving to the left. When a layer of thickness $h$ is turned from the entire surface of the cylinder, the center of gravity will return to its original position. At the beginning of the experiment, $x_{C}=\frac{l}{2}$, the mass of the cylinder $m=\frac{\pi D^{2} l}{4} \rho$, where $\rho$ is the density of the material the cylinder is made of.
After some time, the turning body will consist of two cylinders of different diameters: $D-2 h$ and $D$. The mass of the first cylinder $m_{1}=\frac{\pi(D-2 h)^{2} x}{4} \rho$, the abscissa of the center of gravity of the first cylinder $x_{1}=\frac{x}{2}$. The mass of the second cylinder $m_{2}=$ $=\frac{\pi D^{2}(l-x)}{4} \rho$, the abscissa of the center of gravity of the second cylinder $x_{2}=\frac{l-x}{2}+x=\frac{l+x}{2}$.[^0]
Fig. 20
Then the abscissa of the center of gravity of the system of two cylinders is
$$
\begin{gathered}
x_{C}=\frac{m_{1} x_{1}+m_{2} x_{2}}{m_{1}+m_{2}}= \\
=\frac{\frac{\pi(D-2 h)^{2} x^{2} \rho}{8}+\frac{\pi D^{2}(l-x) \rho(l+x)}{8}}{\frac{\pi(D-2 h)^{2} x \rho}{4}+\frac{\pi D^{2}(l-x) \rho}{4}}= \\
=\frac{(D-2 h)^{2} x^{2}+D^{2}\left(l^{2}-x^{2}\right)}{2\left((D-2 h)^{2} x+D^{2}(l-x)\right)}=\frac{4 h(h-D) x^{2}+D^{2} l^{2}}{8 h(h-D) x+2 D^{2} l}
\end{gathered}
$$
The problem has been reduced to finding the maximum value of the function
$$
y=\frac{4 h(h-D) x^{2}+D^{2} l^{2}}{8 h(h-D) x+2 D^{2} l}
$$
Let's find the range of values of this function:
$E(f)=\left\{y \left\lvert\, y=\frac{4 h(h-D) x^{2}+D^{2} l^{2}}{8 h(h-D) x+2 D^{2} l}\right.\right.$ has a solution in $[0 ; l[\}=$ $=\left\{y \mid 4(h-D) h x^{2}-8 h(h-D) y x+D^{2} l^{2}-2 D^{2} l y=0\right.$ has a solution in $\left[0 ; l[\}=\left\{y \left\lvert\,\left(y \geqslant \frac{D l}{2 h} \cup y \leqslant \frac{D l}{2(D-h)}\right) \cap 0 \leqslant y \leqslant l\right.\right\}=\{y \mid 0 \leqslant\right.$ $\left.\leqslant y \leqslant \frac{D l}{2(D-h)}\right\}$
Let's explain the last conclusion: $D>2 h \Rightarrow \frac{D}{2 h}>1 \Rightarrow \frac{D l}{2 h}>l$. Therefore, we have $0 \leqslant y \leqslant \frac{D l}{2(D-h)}$.
$$
\begin{gathered}
y_{\text {max }}=\frac{D l}{2(D-h)} ; \quad x_{\text {max }}=\frac{D l}{2(D-h)} \\
C C_{\text {max }}^{\prime}=\frac{D l}{2(D-h)}-\frac{l}{2}=\frac{l h}{2(D-h)}
\end{gathered}
$$
A series of problems reduces to finding the maximum or minimum value of a quadratic trinomial.
T h e o r e m. A quadratic trinomial $y=a x^{2}+b x+c$ has a maximum or minimum value, which is taken at $x=-\frac{b}{2 a}$; the value is minimum if $a>0$, and maximum if $a<0$. If $a>0$, the first term $a\left(x+\frac{b}{2 a}\right)^{2}$ cannot be negative; it becomes zero at $x=-\frac{b}{2 a}$. In this case, $y$ takes its minimum value $y=c-\frac{b^{2}}{4 a}$ and the function does not have a maximum value. If $a<0$, then $a\left(x+\frac{b}{2 a}\right)^{2} \leqslant 0$, but at $x=-\frac{b}{2 a}$ it becomes zero. Therefore, $y$ reaches its maximum value $y=c-\frac{b^{2}}{4 a}$. In this case, there is no minimum value.
Corollary. The product of two positive factors, whose sum is constant, reaches its maximum value when these factors are equal ${ }^{1}$.
Proof. Let $p$ be the sum of these two factors. If the first factor is $x$, then the second is $p-x$. The product of the considered factors $y=x(p-x)=-x^{2}+p x$, as follows from the proven theorem, takes its maximum value, equal to $y=\frac{p^{2}}{4}$, at $x=\frac{p}{2}$.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 35. For the fattening of animals, two types of feed I and II are used. Each kilogram of feed I contains 5 units of nutrient $\boldsymbol{A}$ and 2.5 units of nutrient B, while each kilogram of feed II contains 3 units of nutrient A and 3 units of nutrient B. Experimental data has shown that the fattening of animals will be economically beneficial when each animal receives at least 30 units of nutrient $A$ and at least 22.5 units of nutrient B in their daily diet. It is known that the cost of 1 kg of feed I and 1 kg of feed II is the same and equals 1 monetary unit. What should be the daily consumption of each type of feed to minimize the cost of feeding while meeting the above nutritional requirements?
|
## S o l u t i o n.
1. Let's construct a mathematical model of this problem. Let $x$ and $y$ be the number of kilograms of feed of types I and II, respectively, consumed daily. Then the system of constraints is:
$$
\left\{\begin{array}{l}
5 x+3 y \geqslant 30 \quad(a) \\
2.5 x+3 y \geqslant 22.5 \\
x \geqslant 0, y \geqslant 0
\end{array}\right.
$$
50
The objective function to be minimized is $F=x+y$.
2. Let's construct the set of feasible plans (on Fig. 33, this set is shaded). It is clear that $F$ achieves its minimum value at point $M$.
To find its coordinates, we solve the system
$$
\left\{\begin{array}{l}
5 x+3 y=30 \\
2.5 x+3 y=22.5
\end{array}\right.
$$
We get $(3 ; 5) ; F_{\text {min }}=8$ (monetary units).
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 37. On the set of solutions of the system of constraints
$$
\left\{\begin{array}{l}
2-2 x_{1}-x_{2} \geqslant 0 \\
2-x_{1}+x_{2} \geqslant 0 \\
5-x_{1}-x_{2} \geqslant 0 \\
x_{1} \geqslant 0, \quad x_{2} \geqslant 0
\end{array}\right.
$$
find the minimum value of the function $F=x_{2}-x_{1}$.
|
S o l u t i o n.
The set of feasible plans is the polygon $A B C D E$ (Fig. 35). The line $2-$ $-x_{1}-x_{2}=0$ is parallel to the level lines of the function $F=x_{2}$ $-x_{1}$. Therefore, all points on the segment $C D$ give the same minimum value of the objective function $F=x_{2}-x_{1}$ on the set of points of the pentagon $A B C D E$: $F=-2$.
|
-2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 39. Minimize $\boldsymbol{F}=\boldsymbol{x}_{2}-\boldsymbol{x}_{1}$ for non-negative $x_{1}$ and $x_{2}$, satisfying the system of constraints:
$$
\left\{\begin{aligned}
-2 x_{1}+x_{2}+x_{3} & =2 \\
x_{1}-2 x_{2}+x_{4} & =2 \\
x_{1}+x_{2}+x_{5} & =5
\end{aligned}\right.
$$
56
|
## S o l u t i o n.
These constraints can be considered as derived from inequalities, since each of the variables $x_{3}, x_{4}, x_{5}$ appears only in one equation.
1. Write the constraints as equations expressing the basic variables in terms of the non-basic variables:
$$
\left\{\begin{array}{l}
x_{3}=2+2 x_{1}-x_{2} \\
x_{4}=2-x_{1}+2 x_{2} \\
x_{5}=5-x_{1}-x_{2}
\end{array}\right.
$$
The basis B consists of the variables $x_{3}, x_{4}, x_{5}$. It corresponds to the basic non-negative solution $(0 ; 0 ; 2 ; 2 ; 5)$.
Now we need to express $F$ in terms of the non-basic variables. In our specific case, this has already been done.
2. Check if the objective function has reached its minimum value. The coefficient of $x_{1}$ in the expression for $F$ is negative. Therefore, increasing $x_{1}$ will further decrease $F$. However, as the value of $x_{1}$ increases, the values of $x_{4}$ and $x_{5}$ will decrease, and it is necessary to ensure that none of them become negative. Since increasing the value of $x_{1}$ leads to an increase in the value of $x_{3}$, there is no such danger for this variable. From the analysis of the other variables, we find that the value of $x_{1}$ can only be increased to 2. Such an increase gives $x_{4}=0, x_{3}=6, x_{5}=3$.
The new basis B consists of $x_{1}, x_{3}, x_{5}$.
3. To proceed to the next step, express these variables and the objective function in terms of the non-basic variables $x_{2}$ and $x_{4}$. For this, first solve the second equation of the system (2) for the new basic variable $x_{1}$:
$$
x_{1}=2+2 x_{2}-x_{4}
$$
Substituting this expression into the other equations and the objective function gives:
$$
\begin{aligned}
& x_{3}=6+3 x_{2}-2 x_{4} \\
& x_{5}=3-3 x_{2}+x_{4} \\
& F=-2-x_{2}+x_{4}
\end{aligned}
$$
4. We can further decrease the objective function $F$ by increasing the value of $x_{2}$. However, $x_{2}$ can only be increased to 1: this follows from the equation $x_{5}=3-3 x_{2}+x_{4}$. Substituting $x_{2}=1$ into the other equations gives $x_{1}=4, x_{3}=9$. Once again, express the basic variables and $F$ in terms of the non-basic variables:
$$
\begin{aligned}
& \left\{\begin{array}{l}
x_{1}=4-\frac{1}{3} x_{4}-\frac{2}{3} x_{5} \\
x_{2}=1+\frac{1}{3} x_{4}-\frac{1}{3} x_{5} \\
x_{3}=9-x_{4}-x_{5}
\end{array}\right. \\
& F=-3+\frac{2}{3} x_{4}+\frac{1}{3} x_{5}
\end{aligned}
$$
The basis B consists of the variables $x_{1}, x_{2}, x_{3}$.
5. By increasing the values of $x_{4}$ and $x_{5}$, we can no longer achieve a further decrease in $F$. Therefore, we have obtained the optimal solution.
The minimum value of $F$, equal to -3, is achieved at $x_{1}=4$, $x_{2}=1, x_{3}=9$.
Compare the values of the objective function corresponding to different bases:
$$
\begin{gathered}
F_{\mathrm{B}}=5, F_{\mathrm{b}^{\prime}}=-2, F_{\mathrm{E}^{\prime \prime}}=-3 \\
-3<-2<5
\end{gathered}
$$
|
-3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. If a zero is appended to the right of the thought number and the result is subtracted from 143, the result will be three times the thought number. What number was thought of?
|
Solution. Let the number be $x$. We have the equation $143-10 x=3 x$, from which $x=11$. A n s w e r: 11 .
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
41(1283). The path from $A$ to $B$ goes 3 km uphill, 6 km downhill, and 12 km on flat ground. The motorcyclist covered this path in 1 hour and 7 minutes, and the return path in 1 hour and 16 minutes. Find the motorcyclist's speed uphill and downhill, if his speed on flat ground was 18 km/h. (Note that the motorcyclist traveled at constant speeds both uphill and downhill, both on the path $A B$ and on the return path.)
|
Solution. I method. On a flat surface in one direction, the motorcyclist traveled $\frac{2}{3}$ hours $\left(12: 18=\frac{2}{3}\right)$, or 40 minutes. Then, 3 km uphill and 6 km downhill, the motorcyclist traveled 27 minutes, and 6 km uphill and 3 km downhill, the motorcyclist traveled 36 minutes. If we denote the motorcyclist's speed uphill and downhill as $v_{1}$ (km/min) and $v_{2}$ (km/min) respectively, we will have the system of equations:
$$
\left\{\begin{array}{l}
\frac{3}{v_{1}}+\frac{6}{v_{2}}=27 \\
\frac{3}{v_{2}}+\frac{6}{v_{1}}=36
\end{array}\right.
$$
56
from which $v_{1}=\frac{1}{5}$ km/min, or 12 km/h, $v_{2}=\frac{1}{2}$ km/min, or 30 km/h.
II method. The problem can also be solved without setting up a system of equations.
Indeed, if 3 km uphill and 6 km downhill took the motorcyclist 27 minutes, then by doubling the distance, we find that for a distance of 6 km uphill and 12 km downhill, the motorcyclist would need 54 minutes. Since the motorcyclist, according to the problem, spent 36 minutes on 6 km uphill and 3 km downhill, it would take 18 minutes $(54-36=18)$ for 9 km $(12-3=9)$ downhill. Therefore, the motorcyclist's speed downhill is $\frac{9}{18}=\frac{1}{2}$ km/min, or 30 km/h.
Similarly, we find that for a distance of 12 km uphill and 6 km downhill, the motorcyclist would need 72 minutes $(36 \cdot 2=72)$. Then, for a distance of 9 km uphill, it would take 45 minutes $(72-27=45)$. Therefore, the motorcyclist's speed uphill is $\frac{9}{45}=$ $=\frac{1}{5}$ km/min, or 12 km/h.
Answer: 12 km/h, 30 km/h.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
45 (1287). A rider and a pedestrian simultaneously set off from point $A$ to point $B$. The rider, arriving in $B$ 50 minutes earlier than the pedestrian, returned back to $A$. On the return trip, he met the pedestrian 2 kilometers from $B$. The rider spent 1 hour and 40 minutes on the entire journey. Find the distance from $A$ to $B$ and the speed of the rider and the pedestrian.
|
S o l u t i o n. Since the rider spent 1 hour 40 minutes on the entire journey, he spent 50 minutes on the journey from $A$ to $B$ (1 hour 40 minutes: $2=50$ minutes). Then the pedestrian spent 1 hour 40 minutes on the journey from $A$ to $B$, as he arrived in $B$ 50 minutes later than the rider. Therefore, the pedestrian's speed is half the rider's speed. This means that when the rider was in $B$, the pedestrian was halfway through the journey. Since they met 2 km from $B$ on the return journey, the pedestrian had traveled 1 km from the halfway point (2 km $: 2=1$ km). Therefore, the distance from the halfway point to $B$ is 3 km, and the total distance between $A$ and $B$ is 6 km. Since the rider covered this distance in 50 minutes $\left(\frac{5}{6}\right.$ hours $)$, his speed is 7.2 km/h $\left(6: \frac{5}{6}=7.2\right)$; the pedestrian's speed is half of that: $7.2: 2=3.6$ km/h.
A n s w e r: 6 km; 7.2 km/h; 3.6 km/h.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
47 (1289). Two brothers walk together from school to home at the same speed. One day, 15 minutes after leaving school, the first brother ran back to school and, upon reaching it, immediately set off to catch up with the second. Left alone, the second continued walking home at half his usual speed. When the first brother caught up with the second, they resumed their initial speed and arrived home 6 minutes later than usual. How many times faster is the first brother's running speed compared to their usual walking speed?
|
Solution. Let the initial speed of the brothers be denoted by $v$ (m/min), and the segment of the path that the second brother traveled at a speed of $\frac{v}{2}$ be denoted by $s$. For this segment, he spent $\frac{2 s}{v}$ minutes $\left(s: \frac{v}{2}\right)$, which is 6 minutes more than the usual time spent (at speed $v$). We have the equation:
$$
\frac{2 s}{v}-\frac{s}{v}=6 \text {, i.e., } \frac{s}{v}=6, s=6 v \text {. }
$$
The first brother, in the same time $\left(\frac{2 s}{v}=\frac{2 \cdot 6 v}{v}=12\right.$ minutes), ran a distance of $15 v+15 v+s=30 v+6 v=36 v$. His running speed is $36 v: 12=3 v$, i.e., three times the usual walking speed of the brothers.
Answer: three times.
■ 48(1134)'. In a school mathematics olympiad, 9 sixth-grade students participated. For each solved problem, a student received 2 points, and for each unsolved problem, 1 point was deducted. A total of 10 problems were offered. Prove that among the participants from the sixth grade, there were at least 2 students who scored the same number of points. (It is assumed that a student who received more penalty points than earned points scored zero points.)
Solution. To prove this, we will create a table showing the number of points scored depending on the number of solved problems.
| Number of solved problems | 10 | 9 | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 |
| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Number of points scored | 20 | 17 | 14 | 11 | 8 | 5 | 2 | 0 | 0 | 0 |
From the table, it is clear that there are only 8 different possibilities for scoring points. Since there were 9 students, at least two of them must have scored the same number of points.
49(1135). Two tourists, with one bicycle, need to travel a distance of 12 km in 1.5 hours. On the bicycle, each can reach a speed of 20 km/h, and on foot, 5 km/h. Can the tourists complete the entire journey without being late, if they cannot ride the bicycle together? (The bicycle can be left unattended.)
Solution. Let the first tourist ride the bicycle for $x$ km and walk the remaining $12-x$ km. Then the second tourist will ride the bicycle for $12-x$ km and walk $x$ km. Since the tourists will gain the most time if they arrive at the final destination simultaneously, we have the equation:
$$
\frac{x}{20}+\frac{12-x}{5}=\frac{x}{5}+\frac{12-x}{20}, \text { from which } x=6
$$
Thus, the tourists can complete the entire journey without being late if each of them rides the bicycle for half the distance (6 km) and walks the other half. They can switch modes of transportation, for example, every 100 meters.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
105(974). Determine the value of $b$, if the graph of the function defined by the formula passes through the point with coordinates $(3 ; 10)$:
1) $y=x+b$;
2) $y=3 x+b$
3) $y=-\frac{1}{3} x+b$;
4) $y=-\frac{1}{2} x+b$.
A n s w e r: 1) $b=7$; 2) $b=1$; 3) $b=11$; 4) $b=11.5$.
106(975). Define the function by a formula whose graph is a straight line passing through points $A$ and $B$:
1) $A(-6 ;-3), B(2 ;-3)$
2) $A(-4 ;-4), B(3,3)$
3) $A(2 ; 2), B(0 ; 4)$;
4) $A(3 ;-8), B(-5 ; 32)$.
A n s w e r: 1) $y=-3$; 2) $y=x$; 3) $y=-x+4$; 4) $y=-5 x+7$.
N o t e. During club sessions, students can be introduced to the equation of a straight line passing through two points with given coordinates. Otherwise, solving such types of problems reduces to solving a system of two linear equations with two unknowns, which we obtain by substituting the coordinates of the given points into the equation of the line $y=k x+b$. Thus, solving problem (4) we get the system:
$$
\left\{\begin{array}{l}
-8=3 k+b \\
32=-5 k+b
\end{array}\right.
$$
solving which we find $k=-5, b=7$, hence $y=-5 x+7$.
$\triangle$ 107(976). A cylindrical helical spring of height 122 mm is to be made from steel wire with a diameter of 5 mm. Find the number of turns of the spring, if the gap between the turns of the unloaded spring should be 8 mm.
|
Solution. Let $x$ be the number of turns in the spring. Considering that the number of gaps in the spring is one less than the number of turns, according to the problem, we set up the equation $5x + 8(x-1) = 122$, from which $x = 10$.
Answer: 10 turns.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
108(977). A person, living in a cottage near a railway station, usually manages to walk from the cottage to the station in time for the train departure in 18 minutes. Once, this person was delayed at home before leaving by several minutes. Although after that, he walked 1.2 times faster than usual, he still missed the train by 2 minutes. How many minutes was he delayed at home before leaving?
|
Solution. If the usual speed of the gardener is $v$ m/min, then the distance from the garden to the station is $18 v$ meters. When the gardener walked at a speed of $1.2 v$ m/min, it took 15 minutes to travel from home to the garden $(18 v: 1.2 v=15)$. Since the gardener was 2 minutes late, the train left 13 minutes after the gardener left home $(15-2=13)$. Therefore, he was delayed at home for 5 minutes $(18-13=5)$.
Answer: 5 minutes.
$\triangle 109(978)$. At 13:00, water started to be filled into a swimming pool from one pipe to fill it by 16:00 the next day. After some time, another identical pipe was turned on because the pool needed to be filled by 12:00 the next day. At what time was the second pipe turned on?
Solution. Since the pool needed to be filled 4 hours earlier $(16-12=4)$, the second pipe (identical to the first) was turned on 4 hours earlier. Therefore, the second pipe was turned on at 8:00 AM $(12-4=8)$.
Answer: at 8:00 AM.
$\triangle 110(979)$. An electric train passed a signal light in 5 seconds and a platform 150 meters long in 15 seconds. What is the length of the train and its speed?
Solution. Method I. Let the speed of the train be $x$ m/s. Then the length of the train is $5 x$ meters. In 15 seconds, the train travels a distance of $15 x$ meters, or $(150+5 x)$ meters. We have the equation $15 x=150+5 x$, from which $x=15$, i.e., the speed of the train is $15 \mathrm{~m} / \mathrm{s}$, and its length is 75 meters.
Method II. Since the time it takes for any point on the train to pass the platform is 10 seconds $(15-5=10)$, the speed of the train is 15 m/s $(150: 10=15)$, and its length is 75 meters $(15 \cdot 5=75)$.
Answer: $75 \mathrm{~m} ; 15 \mathrm{~m} / \mathrm{s}$.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
120 (989). The road from point $A$ to point $B$ is 11.5 km long and goes first uphill, then on flat ground, and finally downhill. The pedestrian spent 2 hours and 54 minutes on the journey from $A$ to $B$, and 3 hours and 6 minutes on the return journey. The pedestrian's walking speed uphill was 3 km/h, on flat ground - 4 km/h, and downhill - 5 km/h. How many kilometers does the part of the road that goes on flat ground constitute?
|
Solution. Let the length of the road uphill be $x$ kilometers, on the plain be $y$ kilometers, and downhill be $z$ kilometers. According to the problem, we form a system of three linear equations with three unknowns:
88
$$
\left\{\begin{array}{l}
\frac{x}{3}+\frac{y}{4}+\frac{z}{5}=2.9 \\
\frac{z}{3}+\frac{y}{4}+\frac{x}{5}=3.1 \\
x+y+z=11.5
\end{array}\right.
$$
By adding the first two equations of the system, we get the equation
$$
\frac{8}{15}(x+z)+\frac{y}{2}=6 .
$$
Substituting the value of $x+z$ from the third equation of the system into this equation, we get the equation
$$
\frac{8}{15}\left(\frac{23}{2}-y\right)+\frac{y}{2}=6
$$
from which $y=4$.
Remark. The solution to the problem can be reduced to forming and solving a system of two linear equations with two unknowns if the length of the road on the plain is expressed as the difference between the total length of the road (11.5 km) and the lengths of the roads uphill and downhill.
Answer: 4 km.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
139. On a meadow, grass grows. If 9 cows were let onto the meadow, they would empty it in 4 days. If 8 cows were let onto the meadow, they would eat all the grass in 6 days. How many cows can graze on the meadow all the time while the grass grows?
|
Solution. Let $x$ kilograms of grass grow on the meadow per day, and one cow eats $y$ kilograms of grass per day. We have the system of equations:
$$
\left\{\begin{array}{l}
a+4 x-9 \cdot 4 y=0 \\
a+6 x-8 \cdot 6 y=0
\end{array}\right.
$$
where $a$ is the amount of grass that grew on the meadow before the cows were let in. Subtracting the first equation from the second, we find $x=6 y$. If the desired number of cows is $n$, then $n y=x$, from which $n=\frac{x}{y}=6$.
Remark. This problem can also be solved without setting up a system of equations. Indeed, if $a$ kilograms is the initial amount of grass, and $x$ kilograms is the amount of grass that grows on the meadow per day, then in one day one cow eats $\frac{a+4 x}{36}$ or $\frac{a+8 x}{48}$ kilograms of grass. We have the equation
$$
\frac{a+4 x}{36}=\frac{a+8 x}{48},
$$
from which $a=2 x$. Therefore, one cow eats $\frac{x}{6}$ kilograms of grass per day, where $x$ kilograms is the amount of grass that grows on the meadow per day. Therefore, 6 cows can be fed on the meadow.
Answer: 6 cows.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
158. It is known that the number $a$ is $n$ times greater than the number $b$, and the sum of the numbers $a$ and $b$ is $m$ times greater than their difference. Find the sum of the numbers $m$ and $n$, if $m$ and $n$ are natural numbers.
94
|
Given the problem, we set up the system of equations:
$$
\left\{\begin{array}{l}
\frac{a}{b}=n \\
\frac{a+b}{a-b}=m
\end{array}\right.
$$
From the second equation, express $\frac{a}{b}$. We have:
$$
\begin{gathered}
a+b=m a-m b, a(1-m)=-b(m+1) \\
\frac{a}{b}=\frac{m+1}{m-1}=\frac{m-1+2}{m-1}=1+\frac{2}{m-1} .
\end{gathered}
$$
Therefore,
$$
\frac{a}{b}=n=1+\frac{2}{m-1}
$$
Since $m$ and $n$ are natural numbers, then $n=3$ when $m=2$, $n=2$ when $m=3$, and $m+n=5$.
Answer: $m+n=5$.
$\triangle$ 159. Justify the rule for multiplying two-digit numbers by 11.
Solution.
$$
\overline{a b} \cdot 11=(10 a+b)(10+1)=100 a+10(a+b)+b
$$
where $a$ is the hundreds digit, $a+b$ is the tens digit, and $b$ is the units digit.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
173. In the Olympiad, 55 schoolchildren participated. All of them submitted their work. During the checking of each problem, one of three grades was given: «+» - the problem was solved, «-» - the problem was attempted, «0» - the problem was not attempted. After checking all the works, it turned out that in no two works did the number of «+» grades and the number of «-» grades coincide simultaneously. What is the smallest number of problems that could have been offered at the Olympiad?
|
Instruction. If the number of tasks was $a$, then
$$
\frac{(a+1)(a+2)}{2}=55 \text { when } a=9 .
$$
A n s w e r: 9 tasks.
|
9
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
15(1091). Prove that the value of the expression $\sqrt{11+6} \sqrt{ } 2+$ $+\sqrt{11-6 \sqrt{2}}$ is a natural number.
|
Proof. I method. Let
$$
\sqrt{11+6 \sqrt{2}}+\sqrt{11-6 \sqrt{2}}=A
$$
Then
$$
A^{2}=11+6 \sqrt{2}+11-6 \sqrt{2}+2 \sqrt{121-72}=36
$$
from which
$$
\sqrt{A^{2}}=|A|=6, \text { i.e. }|\sqrt{11+6 \sqrt{2}}+\sqrt{11-6 \sqrt{2}}|=6
$$
Since under the modulus sign is the sum of two radicals, each of which is a positive number, then
$$
|\sqrt{11+6 \sqrt{2}}+\sqrt{11-6 \sqrt{2}}|=\sqrt{11+6 \sqrt{2}}+\sqrt{11-6 \sqrt{2}}=6
$$
Since $6 \in \boldsymbol{N}$, the statement is proved.
II method. Since $11+6 \sqrt{2}=9+2 \cdot 3 \sqrt{2}+2=(3+\sqrt{2})^{2}$, $11-6 \sqrt{2}=(3-\sqrt{2})^{2}$, then $\sqrt{11+6 \sqrt{2}}+\sqrt{11-6 \sqrt{2}}=\sqrt{(3+\sqrt{2})^{2}}+$ $+\sqrt{(3-\sqrt{2})^{2}}=|3+\sqrt{2}|+|3-\sqrt{2}|=3+\sqrt{2}+3-\sqrt{2}=6$.
Remark. The equalities $11+6 \sqrt{2}=(3+\sqrt{2})^{2}$ and $11-6 \sqrt{2}=(3-\sqrt{2})^{2}$ can also be obtained using the method of undetermined coefficients. Indeed, if $11+6 \sqrt{2}=(a+b \sqrt{2})^{2}$, where $a \in \boldsymbol{N}$ and $b \in \boldsymbol{N}$, then from the equality $\left(a^{2}+2 b^{2}\right)+2 a b \sqrt{2}=11+6 \sqrt{2}$ we conclude that $a^{2}+2 b^{2}=11, a b=3$, from which $a=3, b=1$, i.e. $11+6 \sqrt{2}=(3+\sqrt{2})^{2}$. Similarly, it is established that $11-$ $-6 \sqrt{2}=(3-\sqrt{2})^{2}$.
|
6
|
Algebra
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
23(1099). A bus left city $M$ for city $N$ at a speed of 40 km/h. A quarter of an hour later, it met a car traveling from city $N$. This car reached city $M$, 15 minutes later it set off back to city $N$ and overtook the bus 20 km from city $N$. Find the distance between cities $M$ and $N$, if the speed of the car is 50 km/h.
|
Solution. Since the bus traveled 10 km in a quarter of an hour $(40: 4=10)$, the bus met the car 10 km from city $M$. If the distance between the cities is $s$ kilometers, then by the time the car overtook the bus, the bus had traveled $(s-30)$ kilometers from the moment of the meeting, spending $\frac{s-30}{40}$ hours on this. By that time, the car had traveled 20 km more, i.e., $(s-10)$ kilometers, spending $\frac{s-10}{50}$ hours on the entire journey (from the moment of the meeting). Considering that the car stood in city $M$ for 15 minutes $\left(\frac{1}{4} \text{ hours}\right)$, we get the equation
$$
\frac{s-10}{50}+\frac{1}{4}=\frac{s-30}{40}
$$
from which $s=160$ km.
Answer: 160 km.
$24(1100)$. Two boys started on a running track 50 m long with a 1-second interval. The boy who started second caught up with the first 10 m from the starting line, ran to the end of the track, and then ran back at the same speed. At what distance from the end of the track did he meet the first boy, if it is known that this meeting occurred 10 seconds after the first boy started?
Solution. Let the speed of the boy who started first be $v_{1}$ meters per second, and the speed of the boy who started second be $v_{2}$ meters per second $\left(v_{2}>v_{1}\right)$. The distance of 10 m, each of the starters ran in $\frac{10}{v_{1}}$ seconds and $\frac{10}{v_{2}}$ seconds, respectively. Since one boy caught up with the other 10 m from the starting line, and they started with a 1-second interval, we have the equation
$$
\frac{10}{v_{1}}-\frac{10}{v_{2}}=1
$$
Since, according to the problem, the boys met 10 seconds after the first boy started (9 seconds after the second boy started) and by the time of the meeting, both had run a distance equal to twice the length of the running track ( $50 \cdot 2=100$ ), we have the equation
$$
10 v_{1}+9 v_{2}=100
$$
Thus, we have a system of two equations with two unknowns:
$$
\left\{\begin{array}{l}
\frac{10}{v_{1}}-\frac{10}{v_{2}}=1 \\
10 v_{1}+9 v_{2}=100
\end{array}\right.
$$
From the first equation, express one of the unknowns (for example, $v_{2}$) and substitute its value into the second equation. We have $v_{2}=\frac{10 v_{1}}{10-v_{1}}, v_{1}^{2}-29 v_{1}+100=0$, from which $v_{1}^{\prime}=4 \text{ m/s}, v_{1}^{\prime \prime}=25 \text{ m/s}$.
Obviously, the condition of the problem is satisfied by one root $v_{1}=$ $=4 \text{ m/s}$.
Thus, the speed of the first boy is $4 \text{ m/s}$. The second boy met the first at a distance of 10 m from the end of the track $(10=50-10 \cdot 4)$.
Remark. To verify, it is advisable to find the speed of the second boy $\left(v_{2}=\frac{10 v_{1}}{10-v_{1}}=\frac{10 \cdot 4}{10-4}=6 \frac{2}{3} \text{ m/s}\right)$ and calculate the required distance in another way:
$$
9 v_{2}-50=9 \cdot \frac{20}{3}-50=10 \text{ m}
$$
Answer: 10 m.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
33(1109). Find the members of the proportion $x_{1}: x_{2}=x_{3}: x_{4}$, where the first term is 6 more than the second, and the third is 5 more than the fourth. The sum of the squares of all terms is 793.
|
Solution. By the condition $x_{1}=x_{2}+6, x_{3}=x_{4}+5,\left(x_{2}+6\right): x_{2}=$ $=\left(x_{4}+5\right): x_{4}$.
We have a system of two equations with two unknowns:
$$
\left\{\begin{array}{l}
\left(x_{2}+6\right)^{2}+x_{2}^{2}+\left(x_{4}+5\right)^{2}+x_{4}^{2}=793 \\
\left(x_{2}+6\right) \cdot x_{4}=x_{2} \cdot\left(x_{4}+5\right)
\end{array}\right.
$$
from which
$$
\left\{\begin{array}{l}
x_{2}^{2}+6 x_{2}+x_{4}^{2}+5 x_{4}-366=0 \\
6 x_{4}=5 x_{2}
\end{array}\right.
$$
Substituting the value $x_{4}=\frac{5 x_{2}}{6}$ from the second equation of the system into the first, we get:
$$
\begin{aligned}
& x_{2}^{2}+6 x_{2}-216=0 \\
& x_{2}^{\prime}=-18, x_{2}^{\prime \prime}=12
\end{aligned}
$$
Since $x_{1}=x_{2}+6$, then $x_{1}^{\prime}=-12, x_{1}^{\prime \prime}=18$.
$$
x_{4}^{\prime}=\frac{5 x_{2}^{\prime}}{6}=-15, x_{4}^{\prime \prime}=\frac{5 x_{2}^{\prime \prime}}{6}=10, x_{3}^{\prime}=x_{4}^{\prime}+5=-10, x_{3}^{\prime \prime}=15
$$
We obtained two proportions:
$$
(-12):(-18)=(-10):(-15) \text { and } 18: 12=15: 10 .
$$
Answer: $18: 12=15: 10 ;(-12):(-18)=(-10):(-15)$.
$\triangle 34$ (1110). From a city, two pedestrians set out at different times along two mutually perpendicular roads. The speed of one of them is 4 km $/ h$, and the other's is 5 km $/ h$. Currently, the pedestrian walking at 4 km $/ h$ is 7 km from the city, and the second is 10 km away. In how many hours will the distance between the pedestrians be 25 km?
Solution. Let the distance of 25 km between the pedestrians be after $t$ hours. Then, based on the Pythagorean theorem, we can form the equation $(7+4 t)^{2}+(10+5 t)^{2}=25^{2}$, from which
$$
41 t^{2}+156 t-476=0, t=2
$$
By checking, we easily confirm that the problem is solved correctly. Indeed, after 2 hours, the first pedestrian was 15 km from the city $(7+2 \cdot 4=15)$, and the second - 20 km from the city $(10+2 \cdot 5=20)$. The distance between the pedestrians $\sqrt{15^{2}+20^{2}}=$ $=\sqrt{625}=25$, which meets the condition of the problem.
Answer: in 2 hours.
3 remark. After this problem, it is advisable to solve problem $69 \$ 8$ (No. 1003 from [2]).
$\triangle 1111$. Prove that if $z=\frac{2}{\frac{1}{a}+\frac{1}{b}}(a \neq 0, \quad b \neq 0$, $a+b \neq 0, a-b \neq 0)$, then $\frac{1}{z-a}+\frac{1}{z-b}=\frac{1}{a}+\frac{1}{b}$.
Proof. Transform the given expression $z$ :
$$
z=\frac{2}{\frac{1}{a}+\frac{1}{b}}=\frac{2 a b}{a+b}
$$
Substitute the value $z=\frac{2 a b}{a+b}$ into the left side of the identity to be proven. We will have:
$$
\begin{gathered}
\frac{1}{z-a}+\frac{1}{z-b}=\frac{1}{\frac{2 a b}{a+b}-a}+\frac{1}{\frac{2 a b}{a+b}-b}= \\
=\frac{a+b}{2 a b-a^{2}-a b}+\frac{a+b}{2 a b-a b-b^{2}}=\frac{a+b}{a b-a^{2}}+\frac{a+b}{a b-b^{2}}= \\
=\frac{a+b}{a(b-a)}+\frac{a+b}{b(a-b)}=\frac{a+b}{a(b-a)}-\frac{a+b}{b(b-a)}=\frac{a b+b^{2}-a^{2}-a b}{a b(b-a)}= \\
=\frac{b^{2}-a^{2}}{a b(b-a)}=\frac{b+a}{a b}=\frac{1}{a}+\frac{1}{b} .
\end{gathered}
$$
We obtained the expression on the right side of the identity. The statement is proven.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
46(1122). Several points are marked on a plane, no three of which lie on the same line. A line is drawn through every two points. How many points are marked on the plane if it is known that a total of 45 lines have been drawn?
|
Solution. The problem reduces to solving the equation $\frac{x(x-1)}{2}=45$ in natural numbers, where $x$ is the number of points marked on the plane. We have $x^{2}-x-90=0$, from which $x_{1}=10, x_{2}=-9$. Only the root $x=10$ satisfies the condition of the problem.
Answer: 10 points.
$\triangle 47(969)$ '. Prove that to find the square of a two-digit number ending in the digit 5 and having $n$ tens, it is sufficient to multiply the number of tens $n$ by $n+1$ and append 25 (see problem 1091 from [4]).
Solution. Let's square the two-digit number $a 5=10 a+5$. We have:
$$
(10 a+5)^{2}=100 a^{2}+100 a+25=100 a(a+1)+25
$$
We obtained that the square of the number $\overline{a 5}$ ends with the number 25, and the number of its hundreds is $a(a+1)$, which is what we needed to prove.
Remark. It is also advisable to solve problem 856 from $[4]$.
' For problems 47-77, the numbers in parentheses indicate the exercise numbers in the textbook [2] 122
$\triangle$ 48(970). Find a two-digit number whose square is written using the digits $0 ; 2 ; 3$ and 5.
$\mathrm{Pe}$ solution. Since the square of an integer cannot end in the digit 2, 3, or a single zero, the square of an integer can only end in the digit 5 from the four given digits. Then the second-to-last digit is 2. Considering that the first digit cannot be zero, we establish that the square of a two-digit number is 3025, and the sought number is 55. (To find the sought number, it is useful to use the previous problem.)
|
10
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
51 (974). What is the last digit of the difference $43^{43}-17^{17}$?
|
Solution. Let's determine the last digits of various powers of the numbers 43 and 17. The number 43 ends with the digit $3, 43^2$ ends with the digit $9, 43^3$ ends with the digit $7, 43^4$ ends with the digit 1. Therefore, the last digits in subsequent powers of the number 43 will repeat: $3, 9, 7, 1$, and so on. Since $43^{43} = 43^{4 \cdot 10 + 3} = (43^4)^{10} \cdot 43^3$, and $(43^4)^{10}$ ends with the digit 1, $43^{43}$, like $43^3$, ends with the digit 7.
Similarly, we establish that the last digits of the powers $17^1, 17^2, 17^3, 17^4$ are $7, 9, 3, 1$, respectively, i.e., the last digits of consecutive powers of the number 17 will also repeat every four powers. Since $17^{17} = 17^{4 \cdot 4 + 1} = (17^4)^4 \cdot 17^1$, the number $17^{17}$, like $17^1$, ends with the digit 7 (just like the number $43^{43}$). Therefore, the difference $43^{43} - 17^{17}$ ends with zero.
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
55 (978). There are 4 balls of different masses. Using a balance scale without weights, how many weighings are needed to arrange these balls in order of decreasing mass?
|
Solution. Let $m_{A}, m_{B}, m_{C}, m_{D}$ be the masses of four different balls. We will use a graph to show how the weighings are performed (the arrow points from the heavier ball to the lighter one).
Suppose, for example, after two weighings, we have $m_{A}>m_{B}$ and $m_{C}>m_{D}$ (Fig. $29, a$ ). Next, we will compare $m_{A}$ and $m_{C}$ (the two heavier balls) and $m_{B}$ and $m_{D}$ (the two lighter balls). In this case, there are two possible scenarios (Fig. $29, b, c$ ).
In the first case (when the smaller of $m_{A}$ and $m_{C}$ is greater than the larger of $m_{B}$ and $m_{D}$, Fig. 29, b), no further weighings are needed: if $m_{A}>m_{C}$, and $m_{C}>m_{D}$, then $m_{A}>m_{D}$; if $m_{C}>m_{D}$, and $m_{D}>m_{B}$, then $m_{C}>m_{B}$ (by the transitive property of inequalities). Therefore, $m_{A}>m_{C}>m_{D}>m_{B}$.
In the second case (when the smaller of $m_{A}$ and $m_{C}$ is greater than the smaller of $m_{D}$ and $m_{B}$, Fig. 29, c), one more weighing is needed to compare the masses of $m_{A}$ and $m_{D}$ ( $m_{C}$ and $m_{D}$ can be easily compared using the transitive property of inequalities). (On the diagram, dashed arrows show the relationships between the masses of the balls established without weighing, based on the transitive property of inequalities.)
Answer: no more than 5 weighings.
|
5
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
69(1003). A cyclist and a motorcyclist are moving towards an intersection along two mutually perpendicular roads. At a certain moment, the cyclist is 8 km away, and the motorcyclist is 15 km away from the intersection. After how many minutes will the distance between them be 5 km, if the cyclist's speed is $\frac{1}{3}$ km/min, and the motorcyclist's speed is 1 km/min?
|
Solution. Let $OA=8$ km, $OB=15$ km (Fig. 30). If the distance of 5 km between the motorcyclist and the cyclist will be in $t$ minutes $\left(A_{1} B_{1}=5\right.$ km), then $A A_{1}=\frac{1}{3} t$ (km), $B B_{1}=t$ (km). Then $O A_{1}=O A-\frac{1}{3} t=$
Fig. 30
$=8-\frac{1}{3} t, O B_{1}=O B-t=15-t$. By the Pythagorean theorem from
$\triangle O A_{1} B_{1}$ we have:
$$
\left(8-\frac{1}{3} t\right)^{2}+(15-t)^{2}=25
$$
After transformations, we get:
$$
\begin{gathered}
5 t^{2}-159 t+1188=0 \\
t_{1}=12\left(\text { min), }, t_{2}=19.8(\text { min }) .\right.
\end{gathered}
$$
Both roots satisfy the condition of the problem: in the first case ($t=12$ min) the motorcyclist and the cyclist have not yet reached the intersection, in the second case ($t=19.8$ min) the motorcyclist will have crossed the intersection.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
73(1007). A master and his apprentice were supposed to complete a job by a certain deadline. However, when half of the work was done, the apprentice fell ill, and the master, left alone, finished the job 2 days late. How many days would it take for each of them to complete the entire job working alone, if the master would need 5 fewer days than the apprentice?
|
Solution. Let the master be able to complete the entire job in $x$ days, then the apprentice - in $(x+5)$ days. In one day, the master completed $\frac{1}{x}$ of the work, the apprentice completed $\frac{1}{x+5}$ of the work; when working together, the master and the apprentice completed $\frac{2 x+5}{x(x+5)}$ of the work per day $\left(\frac{1}{x}+\frac{1}{x+5}=\frac{2 x+5}{x(x+5)}\right)$. For half the work, the master and the apprentice spent $\frac{x(x+5)}{2(2 x+5)}$ days $\left(\frac{1}{2}: \frac{2 x+5}{x(x+5)}\right)$, which is 2 days less than the time spent by the master alone on half the work.
Considering that the master, working alone, spent $\frac{x}{2}$ days on half the work, we set up the equation
$$
\frac{x}{2}-\frac{x(x+5)}{2(2 x+5)}=2
$$
After simplifications, we get:
$$
\begin{gathered}
x^{2}-8 x-20=0 \\
x_{1}=10, x_{2}=-2
\end{gathered}
$$
Only the positive root satisfies the condition of the problem: the master could complete the work in 10 days; therefore, the apprentice in 15 days.
Answer: 10 days, 15 days.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
84. Nine identical books cost 11 rubles and some kopecks, while 13 such books cost 15 rubles and some kopecks. Determine the exact cost of one book
|
S o l u t i o n. It is clear that each book costs 1 r. $m$ k., where $m$ is an integer, and $m<100$. From the first condition of the problem, it follows that $200<9 m<300$, i.e., $23 \leqslant m \leqslant 33$. From the second condition, $200<13 m<300$, i.e., $16 \leqslant m \leqslant 23$. Therefore, $m=23$.
A n s w e r: 1 r. 23 k.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
89. Calculate the value of the expression $\frac{2 a-b}{3 a-b}+\frac{5 b-a}{3 a+b}$, given that $10 a^{2}-3 b^{2}+5 a b=0$ and $9 a^{2}-b^{2} \neq 0$.
|
Solution. Since $5 a b=3 b^{2}-10 a^{2}$, then
$$
\begin{aligned}
& \quad \frac{2 a-b}{3 a-b}+\frac{5 b-a}{3 a+b}=\frac{3 a^{2}+15 a b-6 b^{2}}{9 a^{2}-b^{2}}=\frac{3 a^{2}+3\left(3 b^{2}-10 a^{2}\right)-6 b^{2}}{9 a^{2}-b^{2}}= \\
& \text { Answer: }-3 . \\
& =\frac{-3\left(9 a^{2}-b^{2}\right)}{9 a^{2}-b^{2}}=-3 .
\end{aligned}
$$
|
-3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
96. The number 392 was divided by a natural number $a$, and from the quotient, $a$ was subtracted. The same operation was performed with the resulting difference, and the same operation was performed again with the new result. The final answer was the number $-a$. What is the value of $a$?
|
Solution. According to the problem, we form the equation
$$
\left(\left(\frac{392}{a}-a\right): a-a\right): a-a=-a
$$
from which
$$
\begin{aligned}
\left(\frac{392}{a}-a\right): a-a & =0, \\
a^{3}+a^{2}-392=0, a^{2}(a+1) & =392, \text { where } a \in \boldsymbol{N} .
\end{aligned}
$$
We have that 392 is divisible by the square of a natural number $a$. Since $392=7^{2} \cdot 2^{3}$, 392 is divisible only by the squares of the numbers 2, 7, and 14. It is clear then that $a=7$.
Answer: $a=7$.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
101. Given two quadratic equations $x^{2}-x+m=0, x^{2}-x+$ $+3 m=0, m \neq 0$. Find the value of $m$ for which one of the roots of the second equation is twice the root of the first equation.
|
Solution. Let $x_{0}$ be the root of the first equation, and $2 x_{0}$ be the root of the second. Then the following numerical equalities hold:
$$
\begin{gathered}
x_{0}^{2}-x_{0}+m=0 \\
4 x_{0}^{2}-2 x_{0}+3 m=0
\end{gathered}
$$
Subtracting the second equality from three times the first, we get $-x_{0}^{2}-x_{0}=0$, from which $x_{0}=0$ or $x_{0}=-1$. If $x_{0}=0$, then from the first equality it follows that $m=0$, which contradicts the condition. If $x_{0}=-1$, then we get $m=-2$. (Verification shows that the value $m=-2$ indeed satisfies the condition of the problem.)
Answer: $m=-2$.
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
106. Several chess players participated in a match tournament, where each participant played several games against every other participant. How many rounds did this competition consist of, if a total of 224 games were played?
|
Solution. Let $x$ be the number of participants in the competition, and $y$ be the number of rounds. In one round, each chess player plays $(x-1)$ games, and all participants together play $\frac{x(x-1)}{2}$ games. Therefore, the total number of games played in the competition is $\frac{x(x-1) y}{2}$. Consequently, $x(x-1) y=448$. Listing the divisors of 448, we notice that only two of them (7 and 8) differ by 1. From the obtained equation, it is easy to find $x=8, y=8$.
Answer: The tournament was held in 8 rounds.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
115. Find the minimum value of the expression $\frac{(4+x)(1+x)}{x}$, where $x$ is a positive number.
|
Solution. Since
$$
\frac{(4+x)(1+x)}{x}=\frac{4}{x}+x+5=2\left(\frac{2}{x}+\frac{x}{2}\right)+5
$$
and $\frac{2}{x}+\frac{x}{2} \geqslant 2$ (for $c>0$, the inequality $c+\frac{1}{c} \geqslant 2$ holds, i.e., 2 is the minimum value of the sum of two reciprocals (see 627 from [2])), the expression $2\left(\frac{2}{x}+\frac{x}{2}\right)+5$ takes its minimum value, which is $9(2 \cdot 2+5=9)$.
Answer: 9.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
116. Find the minimum value of the fraction $\frac{x^{2}-3 x+3}{1-x}$, if $x<1$.
|
Solution. I method. The given fraction can be represented as $\frac{x^{2}}{1-x}+3$. For $x<1$, the first term is non-negative and equals 0 only when $x=0$. Since 0 is included in the set of values for $x$, the given fraction achieves its minimum value of 3 when $x=0$.
II method. Representing the given fraction as the sum
$$
1-x+\frac{1}{1-x}+1
$$
and using the fact that the sum of two positive reciprocals is not less than two, we get that the given fraction is not less than 3, i.e., its minimum value is 3.
Answer: 3.
$10^{*}$
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
117. Find the minimum value of the expression
$$
(x-5)(x-1)(x-6)(x-2)+9
$$
|
S o l u t i o n. We have:
$$
\begin{gathered}
y=(x-5)(x-1)(x-6)(x-2)+9=\left(x^{2}-7 x+10\right) \times \\
\times\left(x^{2}-7 x+6\right)+9=\left(x^{2}-7 x\right)^{2}+16\left(x^{2}-7 x\right)+69= \\
=\left(x^{2}-7 x+8\right)^{2}+5
\end{gathered}
$$
From here, $y_{\text {min }}=5$, when $x^{2}-7 x+8=0$, i.e., when
$$
x=\frac{1}{2}(7 \pm \sqrt{17})
$$
A n s w e r: 5 .
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
121. Find the value of the difference
$$
A=\sqrt{|12 \sqrt{5}-29|}-\sqrt{12 \sqrt{5}+29}
$$
|
S o l u t i o n. Noting that $|12 \sqrt{5}-29|=29-12 \sqrt{5}$, we get:
$$
A=\sqrt{29-12 \sqrt{5}}-\sqrt{29+12 \sqrt{5}}
$$
Since
$$
29-12 \sqrt{5}=20-12 \sqrt{5}+9=(2 \sqrt{5}-3)^{2}
$$
and $29+12 \sqrt{5}=(2 \sqrt{5}+3)^{2}$, we obtain that
$$
A=2 \sqrt{5}-3-(2 \sqrt{5}+3)=-6
$$
A n s w e r: -6 .
|
-6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
123. How many right-angled triangles are there with sides of integer lengths, if one of the legs of these triangles is equal to 15?
|
Solution. If $x$ is the hypotenuse and $y$ is the unknown leg of a right triangle, then $x^{2}-y^{2}=15^{2} ;(x-y)(x+y)=$ $=3 \cdot 3 \cdot 5 \cdot 5$
Since $(x-y)$ and $(x+y)$ are natural numbers, and $x+y > x-y$, there are only 4 cases:
$$
\left\{\begin{array} { l }
{ x - y = 1 } \\
{ x + y = 225 }
\end{array} \left\{\begin{array} { l }
{ x - y = 3 } \\
{ x + y = 75 }
\end{array} \quad \left\{\begin{array}{l}
x-y=5 \\
x+y=45
\end{array} ;\left\{\begin{array}{l}
x-y=9 \\
x+y=25
\end{array}\right.\right.\right.\right.
$$
By solving these 4 systems of equations, we find that there are 4 right triangles that satisfy the given property. Their sides are: 1) $15 ; 112 ; 113 ; 2) 15 ; 36 ; 39 ; 3) 15 ; 20$; $25 ; 4) 15 ; 8 ; 17$.
Answer: 4 triangles.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
130. Find the smallest value of the expression $\left|36^{m}-5^{n}\right|$, where $m$ and $n$ are natural numbers.
|
Solution. The difference $36^{m}-5^{n}$ ends with the digit 1, and the difference $5^{n}-36^{m}$ ends with the digit 9. We will prove that $\left|36^{m}-5^{n}\right|$ cannot be equal to either 1 or 9.
From the equation $36^{m}-5^{n}=1$, it follows that $\left(6^{m}-1\right)\left(6^{m}+1\right)=5^{n}$, and then the power of the number 5 is divisible by the number $6^{m}+1$, which ends with the digit 7, which is impossible. From the equation $5^{n}-36^{m}=9$, it follows that $5^{n}$ is divisible by 9, which is incorrect.
Since $36-5^{2}=11$, then 11 is the smallest value of the given expression.
Answer: 11.
|
11
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
131. A natural number $n$ is the product of four consecutive natural numbers, each greater than 5. Determine the maximum number of possible last digits of $n$, given that its last digit is not 0.
|
Solution. Among four consecutive numbers whose product is equal to $n$, there are two even numbers, and therefore the number $n$ is even. Consequently, none of these numbers is divisible by 5 - otherwise, the number $n$ would end in the digit 0. Thus, we have:
$$
\begin{gathered}
n=(5 k+1)(5 k+2)(5 k+3)(5 k+4)=\left(25 k^{2}+25 k+4\right) \times \\
\times\left(25 k^{2}+25 k+6\right)=(25 k(k+1)+4)(25 k(k+1)+6)= \\
=(50 m+4)(50 m+6)
\end{gathered}
$$
where $m=\frac{k(k+1)}{2}$ is an integer. From this, we get:
$$
\begin{gathered}
n=2500 m^{2}+500 m+24=2000 m^{2}+500 m(m+1)+24= \\
=1000\left(2 m^{2}+p\right)+24
\end{gathered}
$$
where $p=\frac{m(m+1)}{2}$ is an integer.
Thus, the last three digits of the number $n$ are determined - they are 024. The fourth digit from the end of the number is not uniquely determined; for example, $6 \cdot 7 \cdot 8 \cdot 9=3024,11 \cdot 12 \cdot 13 \cdot 14=24024$.
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
139. Can you measure out 10 liters of water using two containers of 9 and 11 liters?
|
Solution. Fill the 11-liter vessel (vessel $A$) with water, then use this water to fill the 9-liter vessel (vessel $B$), empty $B$, and transfer the remaining 2 liters from $A$ to $B$. After this, vessel $A$ will be empty, and vessel $B$ will have 2 liters. Repeating these operations three more times, we will end up with an empty vessel $A$ and 8 liters in vessel $B$. Now, fill vessel $A$ from the tap and use it to top up vessel $B$, leaving 10 liters of water in vessel $A$.
|
10
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
140. Find the greatest common divisor of all nine-digit numbers composed of the digits $1,2,3,4,5,6,7,8,9$ (without repetition).
|
Solution. Each of the 9! such numbers is divisible by 9, since the sum of the digits of each number, which is 45, is divisible by 9. We will prove that 9 is the greatest common divisor of these numbers.
Let the greatest common divisor of the considered numbers be greater than 9. Denote it by $d$. Then the difference between any two of the considered numbers should be divisible by $d$. However, the difference between the numbers 123456798 and 123456789, which is 9, is not divisible by $d$. Therefore, 9 is the greatest common divisor of these numbers.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
141. Find the greatest common divisor of all six-digit numbers composed of the digits $1,2,3,4,5,6$ (without repetition).
|
Re shenie. Each of the 6! numbers is divisible by 3 (since the sum of the digits is 21). We will show that 3 is the greatest common divisor of these numbers.
Using the divisibility rules of numbers, it is easy to show that the greatest common divisor of the considered six-digit numbers cannot be equal to $2, 4, 5, 6, 7, 8, 9$, since among these numbers there is at least one number that does not divide by these numbers (none of the considered numbers is divisible by 9). We will prove that the considered numbers do not have a greatest common divisor greater than 9. Indeed, if $d$ is the greatest common divisor of these numbers, greater than 9, then the difference of any two of them is divisible by $d$. However, the difference between the numbers 123465 and 123456, which is 9, is not divisible by $d$. Therefore, the greatest common divisor of the considered numbers is 3.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
144(1172). Two athletes are running on the same closed track. The speed of each is constant, and one takes 5 seconds less than the other to run the entire track. If they start running simultaneously from the same starting line in the same direction, they will be side by side after 30 seconds. How long will it take for them to meet if they start running simultaneously from the same starting line in opposite directions?
|
Solution. Let $s$ (m) be the length of the closed track, $t$ (s) the time it takes for the first athlete to run the entire track. Then the speed of the first athlete is $\frac{s}{t}(\mathrm{~m} / \mathrm{s})$, and the speed of the second athlete is $\frac{s}{t+5}(\mathrm{m} / \mathrm{s})$.
According to the problem, we set up the equation
$$
30 \cdot \frac{s}{t}-30 \cdot \frac{s}{t+5}=s,
$$
from which $t^{2}+5 t-150=0, t=10$ s. Therefore, the speeds of the athletes are $\frac{s}{10} \mathrm{m} / \mathrm{s}$ and $\frac{s}{15} \mathrm{m} / \mathrm{s}$, respectively, and the desired time is $6 \mathrm{s}\left(s:\left(\frac{s}{10}+\frac{s}{15}\right)=6\right)$.
Answer: 6 s.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
145(1173). A pedestrian and a cyclist set off from point $A$ to point $B$ simultaneously. At point $B$, the cyclist turns back and meets the pedestrian 20 minutes after the start of the journey. Without stopping, the cyclist continues to point $A$, turns back, and catches up with the pedestrian 10 minutes after the first meeting. How long will it take the pedestrian to walk from $A$ to $B$?
|
Solution. Let the speed of the pedestrian be $v_{1}$ meters per minute, the speed of the cyclist be $v_{2}$ meters per minute, and $t$ minutes be the time the pedestrian spent from the moment of the second meeting with the cyclist until arriving at point $B$.
According to the problem, we form a system of equations:
$$
\left\{\begin{array}{l}
\frac{20 v_{1}+10 v_{1}+t v_{1}+10 v_{1}+t v_{1}}{v_{2}}=20 \\
\frac{20 v_{1} \cdot 2+10 v_{1}}{v_{2}}=10
\end{array}\right.
$$
156
$$
\left\{\begin{array}{l}
20 v_{2}=40 v_{1}+2 t v_{1} \\
10 v_{2}=50 v_{1}
\end{array}\right.
$$
from which $t=30$.
Therefore, the required time (the time it takes for the pedestrian to travel from $A$ to $B$) is 1 hour $(20+10+30=60$ (min)).
Answer: 1 hour.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
146(1174). A motorcyclist left point $A$ for point $B$ and at the same time a cyclist left point $B$ for point $A$. The motorcyclist arrived at point $B$ 2 hours after meeting the cyclist, while the cyclist arrived at point $A$ 4.5 hours after meeting the motorcyclist. How many hours were the motorcyclist and the cyclist on the road
|
Solution. Let the motorcyclist and the cyclist have been on the road for $t$ hours before meeting. If $v_{1}$ kilometers per hour is the speed of the motorcyclist, and $v_{2}$ kilometers per hour is the speed of the cyclist, then we have the system of equations:
$$
\left\{\begin{array}{l}
v_{1}=\frac{v_{2} t}{2} \\
v_{2}=\frac{v_{1} t}{4.5}
\end{array}\right.
$$
from which $t^{2}=9, t=3$.
Then the time of the motorcyclist's movement is 5 hours $(3+2=5)$, and the time of the cyclist's movement is 7.5 hours $(3+4.5=7.5)$.
Answer: 5 hours; 7.5 hours.
## § 9. PROBLEMS OF INCREASED DIFFICULTY IN THE ALGEBRA COURSE FOR THE 9TH GRADE
$\triangle 1(1129)^{\prime}$. Find the roots of the polynomial $2 x^{5}+x^{4}-10 x^{3}-$ $-5 x^{2}+8 x+4$. (A root of a polynomial in one variable is a value of the variable for which the value of the polynomial is zero.)
Solution. Factorize the given polynomial:
$$
2 x^{5}+x^{4}-10 x^{3}-5 x^{2}+8 x+4=
$$
$$
=x^{4}(2 x+1)-5 x^{2}(2 x+1)+4(2 x+1)=(2 x+1)\left(x^{4}-5 x^{2}+4\right) .
$$
Set each factor equal to zero. We have $2 x+1=0$, from which $x_{1}=-\frac{1}{2}$, or $x^{4}-5 x^{2}+4=0$, from which $x^{2}=1, x^{2}=4$, i.e., $x_{2}=1, x_{3}=-1, x_{4}=2, x_{5}=-2$.
Answer: $-\frac{1}{2} ;-1 ;-2 ; 1 ; 2$.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4(1132). For what values of $a$ do the quadratic trinomials $x^{2}+$ $+a x+1$ and $x^{2}+x+a$ have a common root?
|
Solution. Let $x=x_{1}$ be the common root of the given quadratic trinomials. Then $x_{1}^{2}+a x_{1}+1=0$ and $x_{1}^{2}+x_{1}+a=0$, i.e., $x_{1}^{2}+a x_{1}+1=x_{1}^{2}+x_{1}+a, \quad a \cdot\left(x_{1}-1\right)=x_{1}-1, \quad\left(x_{1}-1\right) \cdot(a-1)=0$. From this, $a=1$ or $x_{1}=1$.
When $a=1$, each of the given quadratic trinomials has the form $x^{2}+x+1$, which has no solutions in the set of real numbers. Therefore, the given quadratic trinomials have a common root $x=1$. Find the value of $a$ for which $x=1$. When $x=1$, we have $1+a+1=0$, from which $a=-2$.
Answer: $a=-2$.
Remark. It is advisable to do a check. When $a=-2$, we have the quadratic trinomials $x^{2}-2 x+1$ and $x^{2}+x-2$, which have a common root $x=1$.
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5 (1133). For what value of $a$ does the sum of the squares of the roots of the quadratic trinomial $x^{2}-(a-2) x-a-1$ take the smallest value?
|
Solution. Based on Vieta's theorem, we have $x_{1}+x_{2}=$ $=a-2, x_{1} \cdot x_{2}=-a-1$. Then
$$
\begin{gathered}
x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=(a-2)^{2}-2(-a-1)= \\
=a^{2}-2 a+6=(a-1)^{2}+5
\end{gathered}
$$
Since $(a-1)^{2} \geqslant 0$, $x_{1}^{2}+x_{2}^{2}$ takes its minimum value when $a=1$.
Remark 1. This problem can also be solved without using Vieta's theorem, by calculating the roots of the given quadratic trinomial using formulas. We have
158
$$
\begin{gathered}
x_{1}=\frac{a-2+\sqrt{ } a^{2}+8}{2}, x_{2}=\frac{a-2-\sqrt{ } a^{2}+8}{2}, \\
x_{1}^{2}+x_{2}^{2}=\left(\frac{a-2+\sqrt{ } a^{2}+8}{2}\right)^{2}+\left(\frac{a-2-\sqrt{ } a^{2}+8}{2}\right)^{2}= \\
=\frac{4 a^{2}-8 a+24}{4}=a^{2}-2 a+6=(a-1)^{2}+5 .
\end{gathered}
$$
Remark 2. The value of $a$ at which the quadratic trinomial $a^{2}-2 a+6$ obtained during the solution process takes its minimum value can also be found differently, by calculating the x-coordinate of the vertex of the corresponding parabola using the known formula. In our case, $a=\frac{2}{2}=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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